QUESTION 01
Electric Current & Charge
CBSE Board Core
A conductor carries a charge of \( Q = 1.6 \text{ C} \) passing through a cross-section in \( t = 4 \text{ s} \). Calculate:
(a) The electric current (\(I\)) in the conductor.
(b) The total number of electrons (\(n\)) passing through the cross-section (Given \(e = 1.6 \times 10^{-19}\text{ C}\)).
(a) The electric current (\(I\)) in the conductor.
(b) The total number of electrons (\(n\)) passing through the cross-section (Given \(e = 1.6 \times 10^{-19}\text{ C}\)).
Given: \( Q = 1.6\text{ C} \), \( t = 4\text{ s} \), \( e = 1.6 \times 10^{-19}\text{ C} \)
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Detailed Solution 01
(a) Electric Current (\(I\)):
\( I = \frac{Q}{t} = \frac{1.6}{4} = 0.4\text{ A} = 400\text{ mA} \)
(b) Number of Electrons (\(n\)):
\( Q = ne \implies n = \frac{Q}{e} = \frac{1.6}{1.6 \times 10^{-19}} = 1.0 \times 10^{19}\text{ electrons} \)
\( I = \frac{Q}{t} = \frac{1.6}{4} = 0.4\text{ A} = 400\text{ mA} \)
(b) Number of Electrons (\(n\)):
\( Q = ne \implies n = \frac{Q}{e} = \frac{1.6}{1.6 \times 10^{-19}} = 1.0 \times 10^{19}\text{ electrons} \)
QUESTION 02
Current & Milliamperes
CBSE Board Standard
A steady current of \( I = 250 \text{ mA} \) flows through a wire for a duration of \( t = 2 \text{ minutes} \). Calculate:
(a) Total electric charge (\(Q\)) passing through any cross-section of the wire.
(b) Total number of electrons flowing through it.
(a) Total electric charge (\(Q\)) passing through any cross-section of the wire.
(b) Total number of electrons flowing through it.
Given: \( I = 250\text{ mA} = 0.25\text{ A} \), \( t = 2\text{ min} = 120\text{ s} \), \( e = 1.6 \times 10^{-19}\text{ C} \)
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Detailed Solution 02
(a) Total Charge (\(Q\)):
\( Q = I \times t = 0.25\text{ A} \times 120\text{ s} = 30\text{ Coulombs} \)
(b) Number of Electrons (\(n\)):
\( n = \frac{Q}{e} = \frac{30}{1.6 \times 10^{-19}} = 1.875 \times 10^{20}\text{ electrons} \)
\( Q = I \times t = 0.25\text{ A} \times 120\text{ s} = 30\text{ Coulombs} \)
(b) Number of Electrons (\(n\)):
\( n = \frac{Q}{e} = \frac{30}{1.6 \times 10^{-19}} = 1.875 \times 10^{20}\text{ electrons} \)
QUESTION 03Electron Flow DirectionCBSE Conceptual
A beam of \( n = 10^{20} \) electrons moves from terminal A to terminal B in \( t = 0.1 \text{ s} \).
(a) Calculate the magnitude of electric current (\(I\)).
(b) State the direction of conventional electric current flow.
(a) Calculate the magnitude of electric current (\(I\)).
(b) State the direction of conventional electric current flow.
Given: \( n = 10^{20} \), \( t = 0.1\text{ s} \), \( e = 1.6 \times 10^{-19}\text{ C} \)
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Detailed Solution 03
(a) Electric Current (\(I\)):
\( Q = n e = 10^{20} \times 1.6 \times 10^{-19} = 16\text{ C} \)
\( I = \frac{Q}{t} = \frac{16}{0.1} = 160\text{ Amperes} \)
(b) Direction of Conventional Current:
Electrons move from A to B. Conventional current flows in the opposite direction of electron flow, i.e., from B to A.
\( Q = n e = 10^{20} \times 1.6 \times 10^{-19} = 16\text{ C} \)
\( I = \frac{Q}{t} = \frac{16}{0.1} = 160\text{ Amperes} \)
(b) Direction of Conventional Current:
Electrons move from A to B. Conventional current flows in the opposite direction of electron flow, i.e., from B to A.
QUESTION 04Galvanometer & MicroamperesCBSE Numerical
A sensitive galvanometer detects a current of \( I = 50 \,\mu\text{A} \) flowing for \( t = 10 \text{ s} \).
(a) Convert current into amperes.
(b) Calculate the electric charge passing through the galvanometer in this time interval.
(a) Convert current into amperes.
(b) Calculate the electric charge passing through the galvanometer in this time interval.
Given: \( I = 50\,\mu\text{A} = 50 \times 10^{-6}\text{ A} \), \( t = 10\text{ s} \)
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Detailed Solution 04
(a) Conversion:
\( 1\,\mu\text{A} = 10^{-6}\text{ A} \implies I = 50 \times 10^{-6}\text{ A} = 5.0 \times 10^{-5}\text{ A} \)
(b) Charge (\(Q\)):
\( Q = I \times t = (5.0 \times 10^{-5}\text{ A}) \times (10\text{ s}) = 5.0 \times 10^{-4}\text{ Coulombs} \)
\( 1\,\mu\text{A} = 10^{-6}\text{ A} \implies I = 50 \times 10^{-6}\text{ A} = 5.0 \times 10^{-5}\text{ A} \)
(b) Charge (\(Q\)):
\( Q = I \times t = (5.0 \times 10^{-5}\text{ A}) \times (10\text{ s}) = 5.0 \times 10^{-4}\text{ Coulombs} \)
QUESTION 05Charge Flow in LampCBSE Board Core
A total charge of \( Q = 300 \text{ Coulombs} \) flows through the filament of an electric bulb in \( t = 5 \text{ minutes} \). Calculate the current (\(I\)) drawn by the filament.
Given: \( Q = 300\text{ C} \), \( t = 5\text{ min} = 300\text{ s} \)
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Detailed Solution 05
Current (\(I\)):
\( I = \frac{Q}{t} = \frac{300\text{ C}}{300\text{ s}} = 1.0\text{ Ampere} \)
\( I = \frac{Q}{t} = \frac{300\text{ C}}{300\text{ s}} = 1.0\text{ Ampere} \)
QUESTION 06Electric Potential DefinitionCBSE Conceptual
Work done in bringing a charge of \( Q = 5 \text{ C} \) from infinity to a point A in an electric field is \( W = 100 \text{ J} \). Calculate the electric potential (\(V_A\)) at point A.
Given: \( Q = 5\text{ C} \), \( W = 100\text{ J} \)
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Detailed Solution 06
Electric Potential (\(V_A\)):
\( V_A = \frac{W}{Q} = \frac{100\text{ J}}{5\text{ C}} = 20\text{ Volts} \)
\( V_A = \frac{W}{Q} = \frac{100\text{ J}}{5\text{ C}} = 20\text{ Volts} \)
QUESTION 07Potential Difference V_ABCBSE Board Standard
How much work (\(W\)) is done in moving a charge of \( Q = 12.5 \text{ C} \) across two points having a potential difference of \( V = 16 \text{ V} \)?
Given: \( Q = 12.5\text{ C} \), \( V = 16\text{ V} \)
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Detailed Solution 07
Work Done (\(W\)):
\( V = \frac{W}{Q} \implies W = V \times Q = 16\text{ V} \times 12.5\text{ C} = 200\text{ Joules} \)
\( V = \frac{W}{Q} \implies W = V \times Q = 16\text{ V} \times 12.5\text{ C} = 200\text{ Joules} \)
QUESTION 08Voltmeter Connection & WorkCBSE Board Core
(a) Why is a voltmeter always connected in parallel across a resistor?
(b) Calculate the work done in moving \( Q = 4 \text{ C} \) of charge through a potential difference of \( V = 12 \text{ V} \).
(b) Calculate the work done in moving \( Q = 4 \text{ C} \) of charge through a potential difference of \( V = 12 \text{ V} \).
Given: \( Q = 4\text{ C} \), \( V = 12\text{ V} \)
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Detailed Solution 08
(a) Voltmeter Connection:
A voltmeter has very high resistance so that it draws negligible current from the main circuit, ensuring accurate measurement of potential difference across components.
(b) Work Done (\(W\)):
\( W = V \times Q = 12\text{ V} \times 4\text{ C} = 48\text{ Joules} \)
A voltmeter has very high resistance so that it draws negligible current from the main circuit, ensuring accurate measurement of potential difference across components.
(b) Work Done (\(W\)):
\( W = V \times Q = 12\text{ V} \times 4\text{ C} = 48\text{ Joules} \)
QUESTION 09Energy from 6V BatteryCBSE NCERT Direct
How much energy (\(E\)) is given to each coulomb of charge passing through a \( 6 \text{ V} \) battery?
Given: \( Q = 1\text{ C} \), \( V = 6\text{ V} \)
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Detailed Solution 09
Energy Given (\(E = W\)):
\( W = V \times Q = 6\text{ V} \times 1\text{ C} = 6\text{ Joules} \)
Each coulomb of charge gains 6 Joules of electrical energy.
\( W = V \times Q = 6\text{ V} \times 1\text{ C} = 6\text{ Joules} \)
Each coulomb of charge gains 6 Joules of electrical energy.
QUESTION 10Ohm's Law & Circuit CurrentCBSE Board Core
State Ohm's Law. A conductor of resistance \( R = 15 \,\Omega \) is connected across a battery of voltage \( V = 4.5 \text{ V} \). Calculate the current (\(I\)) flowing through the circuit.
Given: \( R = 15\,\Omega \), \( V = 4.5\text{ V} \)
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Detailed Solution 10
Statement: At constant temperature, the current (\(I\)) flowing through a conductor is directly proportional to potential difference (\(V\)) across its ends.
\( V = IR \)
Current Calculation:
\( I = \frac{V}{R} = \frac{4.5\text{ V}}{15\,\Omega} = 0.3\text{ Amperes} = 300\text{ mA} \)
\( V = IR \)
Current Calculation:
\( I = \frac{V}{R} = \frac{4.5\text{ V}}{15\,\Omega} = 0.3\text{ Amperes} = 300\text{ mA} \)
QUESTION 11V-I Graph & ResistanceCBSE Graphical Analysis
A linear \( V-I \) graph for a metallic wire passes through the origin and point \( (I = 0.5\text{ A}, V = 2.0\text{ V}) \).
(a) What physical quantity is represented by the slope of a \( V-I \) graph?
(b) Calculate the resistance (\(R\)) of the wire.
(a) What physical quantity is represented by the slope of a \( V-I \) graph?
(b) Calculate the resistance (\(R\)) of the wire.
Given: \( I = 0.5\text{ A} \), \( V = 2.0\text{ V} \)
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Detailed Solution 11
(a) Slope Significance:
The slope of a \(V-I\) graph (with Voltage on Y-axis and Current on X-axis) represents the Resistance (\(R\)) of the wire.
(b) Resistance (\(R\)):
\( R = \text{Slope} = \frac{\Delta V}{\Delta I} = \frac{2.0\text{ V}}{0.5\text{ A}} = 4.0\,\Omega \)
The slope of a \(V-I\) graph (with Voltage on Y-axis and Current on X-axis) represents the Resistance (\(R\)) of the wire.
(b) Resistance (\(R\)):
\( R = \text{Slope} = \frac{\Delta V}{\Delta I} = \frac{2.0\text{ V}}{0.5\text{ A}} = 4.0\,\Omega \)
QUESTION 12V-I Slopes ComparisonCBSE HOTS
Two conductors A and B have \( V-I \) graph lines making angles of \( 60^\circ \) and \( 30^\circ \) with the current axis (X-axis). Which conductor has higher resistance? Calculate the ratio \( R_A / R_B \).
Given: \( \theta_A = 60^\circ \), \( \theta_B = 30^\circ \)
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Detailed Solution 12
Higher Resistance: Conductor A (larger slope \(\tan\theta\)).
Ratio Calculation:
\( R_A = \tan 60^\circ = \sqrt{3} \)
\( R_B = \tan 30^\circ = \frac{1}{\sqrt{3}} \)
\( \frac{R_A}{R_B} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 : 1 \)
Ratio Calculation:
\( R_A = \tan 60^\circ = \sqrt{3} \)
\( R_B = \tan 30^\circ = \frac{1}{\sqrt{3}} \)
\( \frac{R_A}{R_B} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 : 1 \)
QUESTION 13Ohm's Law CalculationCBSE Board Standard
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Given: \( V = 12\text{ V} \), \( I = 2.5\text{ mA} = 2.5 \times 10^{-3}\text{ A} \)
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Detailed Solution 13
Resistance (\(R\)):
\( R = \frac{V}{I} = \frac{12\text{ V}}{2.5 \times 10^{-3}\text{ A}} = 4800\,\Omega = 4.8\text{ k}\Omega \)
\( R = \frac{V}{I} = \frac{12\text{ V}}{2.5 \times 10^{-3}\text{ A}} = 4800\,\Omega = 4.8\text{ k}\Omega \)
QUESTION 14Torch Bulb ResistanceCBSE Board Core
A torch bulb is rated \( 2.5 \text{ V} \) and \( 750 \text{ mA} \). Calculate:
(a) Its resistance when glowing.
(b) Its power rating.
(a) Its resistance when glowing.
(b) Its power rating.
Given: \( V = 2.5\text{ V} \), \( I = 750\text{ mA} = 0.75\text{ A} \)
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Detailed Solution 14
(a) Resistance (\(R\)):
\( R = \frac{V}{I} = \frac{2.5\text{ V}}{0.75\text{ A}} = 3.33\,\Omega \)
(b) Power (\(P\)):
\( P = V \times I = 2.5\text{ V} \times 0.75\text{ A} = 1.875\text{ Watts} \)
\( R = \frac{V}{I} = \frac{2.5\text{ V}}{0.75\text{ A}} = 3.33\,\Omega \)
(b) Power (\(P\)):
\( P = V \times I = 2.5\text{ V} \times 0.75\text{ A} = 1.875\text{ Watts} \)
QUESTION 15Temperature DependenceCBSE Conceptual
How does the resistance of (a) a metallic conductor, and (b) a semiconductor change with an increase in temperature? Give reasons.
Concept: Temperature effect on electrical resistance.
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Detailed Solution 15
(a) Metallic Conductor: Resistance increases with temperature because thermal vibrations of metal ions increase electron collision frequency.
(b) Semiconductor: Resistance decreases with temperature because more covalent bonds break, generating free charge carriers.
(b) Semiconductor: Resistance decreases with temperature because more covalent bonds break, generating free charge carriers.
QUESTION 16Factors Affecting ResistanceCBSE Core Theory
List four factors on which the resistance of a cylindrical conductor depends. Write the mathematical relation defining resistivity (\(\rho\)).
Formula: \( R = \rho \frac{L}{A} \)
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Detailed Solution 16
Factors:
1. Length (\(L\)): \( R \propto L \)
2. Area of cross-section (\(A\)): \( R \propto 1/A \)
3. Nature of material (\(\rho\))
4. Temperature
Resistivity Formula:
\( \rho = \frac{R A}{L} \) (SI Unit: \(\Omega \cdot \text{m}\))
1. Length (\(L\)): \( R \propto L \)
2. Area of cross-section (\(A\)): \( R \propto 1/A \)
3. Nature of material (\(\rho\))
4. Temperature
Resistivity Formula:
\( \rho = \frac{R A}{L} \) (SI Unit: \(\Omega \cdot \text{m}\))
QUESTION 17Wire Stretching ProblemCBSE High Yield
A wire of initial resistance \( R = 6 \,\Omega \) is stretched uniformly to double its original length. Calculate its new resistance (\(R'\)).
Given: \( R = 6\,\Omega \), \( L' = 2L \), Volume constant \( V = A L = A' L' \)
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Detailed Solution 17
Volume Conservation:
\( A' (2L) = A L \implies A' = \frac{A}{2} \)
New Resistance (\(R'\)):
\( R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4 \left(\rho \frac{L}{A}\right) = 4 R = 4 \times 6\,\Omega = 24\,\Omega \)
\( A' (2L) = A L \implies A' = \frac{A}{2} \)
New Resistance (\(R'\)):
\( R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4 \left(\rho \frac{L}{A}\right) = 4 R = 4 \times 6\,\Omega = 24\,\Omega \)
QUESTION 18Radius Halved ProblemCBSE HOTS Numerical
A wire of resistance \( R = 10 \,\Omega \) is drawn out so that its radius is reduced to half (\(r' = r/2\)). Find its new resistance.
Given: \( R = 10\,\Omega \), \( r' = r/2 \implies A' = A/4 \)
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Detailed Solution 18
Volume Conservation:
\( A L = A' L' \implies \pi r^2 L = \pi (r/2)^2 L' \implies L' = 16 L \)
New Resistance (\(R'\)):
\( R' = \rho \frac{L'}{A'} = \rho \frac{16 L}{A/4} = 16 R = 160\,\Omega \)
\( A L = A' L' \implies \pi r^2 L = \pi (r/2)^2 L' \implies L' = 16 L \)
New Resistance (\(R'\)):
\( R' = \rho \frac{L'}{A'} = \rho \frac{16 L}{A/4} = 16 R = 160\,\Omega \)
QUESTION 19Cut & Parallel WireCBSE Board Standard
A wire of resistance \( R = 16 \,\Omega \) is cut into 4 equal parts. These 4 parts are then connected in parallel. Calculate equivalent resistance (\(R_p\)).
Given: \( R = 16\,\Omega \), 4 equal parts \(\implies R_{each} = 4\,\Omega\)
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Detailed Solution 19
Each Part Resistance:
\( r = \frac{16}{4} = 4\,\Omega \)
Parallel Combination (\(R_p\)):
\( \frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1 \implies R_p = 1.0\,\Omega \)
\( r = \frac{16}{4} = 4\,\Omega \)
Parallel Combination (\(R_p\)):
\( \frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1 \implies R_p = 1.0\,\Omega \)
QUESTION 20Resistivity PropertiesCBSE Conceptual
Why are copper and aluminium wires usually employed for electricity transmission, while Nichrome wire is used in electrical heating devices?
Concept: Electrical transmission vs heating element properties.
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Detailed Solution 20
Transmission Wires (Cu/Al): Very low resistivity (\(\sim 10^{-8}\,\Omega\cdot\text{m}\)), minimizing Joule heating energy losses (\(I^2 R\)).
Heating Element (Nichrome): High resistivity (\(\sim 10^{-6}\,\Omega\cdot\text{m}\)) and high melting point; does not oxidize at red-hot temperatures.
Heating Element (Nichrome): High resistivity (\(\sim 10^{-6}\,\Omega\cdot\text{m}\)) and high melting point; does not oxidize at red-hot temperatures.
QUESTION 21Resistivity FormulaCBSE Board Numerical
Copper wire has diameter \( d = 0.5 \text{ mm} \) and resistivity \( \rho = 1.6 \times 10^{-8} \,\Omega\cdot\text{m} \). Find length (\(L\)) to make its resistance \( R = 10 \,\Omega \).
Given: \( d = 0.5\text{ mm} = 5 \times 10^{-4}\text{ m} \), \( r = 2.5 \times 10^{-4}\text{ m} \), \( \rho = 1.6 \times 10^{-8}\,\Omega\text{m} \)
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Detailed Solution 21
Area (\(A\)):
\( A = \pi r^2 = 3.14 \times (2.5 \times 10^{-4})^2 = 1.9625 \times 10^{-7}\text{ m}^2 \)
Length (\(L\)):
\( L = \frac{R A}{\rho} = \frac{10 \times 1.9625 \times 10^{-7}}{1.6 \times 10^{-8}} = 122.65\text{ metres} \)
\( A = \pi r^2 = 3.14 \times (2.5 \times 10^{-4})^2 = 1.9625 \times 10^{-7}\text{ m}^2 \)
Length (\(L\)):
\( L = \frac{R A}{\rho} = \frac{10 \times 1.9625 \times 10^{-7}}{1.6 \times 10^{-8}} = 122.65\text{ metres} \)
QUESTION 22Ratio AnalysisCBSE Board Core
Two wires of same material have lengths in ratio 1:2 and diameters in ratio 2:1. Find the ratio of their electrical resistances (\(R_1 / R_2\)).
Given: \( L_1/L_2 = 1/2 \), \( d_1/d_2 = 2/1 \implies A_1/A_2 = 4/1 \)
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Detailed Solution 22
Resistance Ratio:
\( \frac{R_1}{R_2} = \left(\frac{L_1}{L_2}\right) \times \left(\frac{A_2}{A_1}\right) = \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) = \frac{1}{8} \implies 1 : 8 \)
\( \frac{R_1}{R_2} = \left(\frac{L_1}{L_2}\right) \times \left(\frac{A_2}{A_1}\right) = \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) = \frac{1}{8} \implies 1 : 8 \)
QUESTION 23Alloy PropertiesCBSE Conceptual
Why are alloys commonly used in electrical heating appliances rather than pure metals? State two reasons.
Concept: Alloys vs pure metals in thermal devices.
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Detailed Solution 23
Reasons:
1. Higher Resistivity: Alloys produce more heat energy per unit current.
2. High Oxidation Resistance: Alloys do not burn/oxidize easily at high operating temperatures.
1. Higher Resistivity: Alloys produce more heat energy per unit current.
2. High Oxidation Resistance: Alloys do not burn/oxidize easily at high operating temperatures.
QUESTION 24Series Circuit FundamentalsCBSE Board Core
Three resistors of \( 2\,\Omega, 3\,\Omega, \) and \( 5\,\Omega \) are connected in series across a 10 V battery.
(a) Find equivalent series resistance (\(R_s\)).
(b) Calculate total circuit current (\(I\)).
(c) Calculate voltage drop across the \( 5\,\Omega \) resistor.
(a) Find equivalent series resistance (\(R_s\)).
(b) Calculate total circuit current (\(I\)).
(c) Calculate voltage drop across the \( 5\,\Omega \) resistor.
Given: \( R_1 = 2\,\Omega, R_2 = 3\,\Omega, R_3 = 5\,\Omega, V = 10\text{ V} \)
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Detailed Solution 24
(a) Series Resistance (\(R_s\)):
\( R_s = 2 + 3 + 5 = 10\,\Omega \)
(b) Circuit Current (\(I\)):
\( I = \frac{V}{R_s} = \frac{10\text{ V}}{10\,\Omega} = 1.0\text{ A} \)
(c) Voltage Drop (\(V_3\)):
\( V_3 = I \times R_3 = 1.0\text{ A} \times 5\,\Omega = 5.0\text{ Volts} \)
\( R_s = 2 + 3 + 5 = 10\,\Omega \)
(b) Circuit Current (\(I\)):
\( I = \frac{V}{R_s} = \frac{10\text{ V}}{10\,\Omega} = 1.0\text{ A} \)
(c) Voltage Drop (\(V_3\)):
\( V_3 = I \times R_3 = 1.0\text{ A} \times 5\,\Omega = 5.0\text{ Volts} \)
QUESTION 25Series Ammeter & VoltmeterCBSE NCERT Problem
An ammeter and a voltmeter are connected in a series circuit containing a 6 V battery and two resistors \( 5\,\Omega \) and \( 8\,\Omega \). What will be the readings of the ammeter and the voltmeter across the \( 8\,\Omega \) resistor?
Given: \( V = 6\text{ V} \), \( R_1 = 5\,\Omega \), \( R_2 = 8\,\Omega \)
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Detailed Solution 25
Ammeter Reading (\(I\)):
\( R_s = 5 + 8 = 13\,\Omega \)
\( I = \frac{6\text{ V}}{13\,\Omega} = 0.46\text{ A} \)
Voltmeter Reading (\(V_2\)):
\( V_2 = I \times 8 = 0.46 \times 8 = 3.69\text{ Volts} \)
\( R_s = 5 + 8 = 13\,\Omega \)
\( I = \frac{6\text{ V}}{13\,\Omega} = 0.46\text{ A} \)
Voltmeter Reading (\(V_2\)):
\( V_2 = I \times 8 = 0.46 \times 8 = 3.69\text{ Volts} \)
QUESTION 26Series CombinationCBSE Board Core
Two resistors of \( 4\,\Omega \) and \( 6\,\Omega \) are connected in series to a cell of EMF \( 12\text{ V} \) with internal resistance \( 2\,\Omega \). Calculate total circuit resistance and current.
Given: \( R_1 = 4\,\Omega, R_2 = 6\,\Omega, r = 2\,\Omega, E = 12\text{ V} \)
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Detailed Solution 26
Total Resistance (\(R_{tot}\)):
\( R_{tot} = 4 + 6 + 2 = 12\,\Omega \)
Circuit Current (\(I\)):
\( I = \frac{E}{R_{tot}} = \frac{12\text{ V}}{12\,\Omega} = 1.0\text{ A} \)
\( R_{tot} = 4 + 6 + 2 = 12\,\Omega \)
Circuit Current (\(I\)):
\( I = \frac{E}{R_{tot}} = \frac{12\text{ V}}{12\,\Omega} = 1.0\text{ A} \)
QUESTION 27Equal Series ResistorsCBSE Numerical
Three identical resistors each of resistance \( R \) are connected in series. If their total equivalent resistance is \( 45 \,\Omega \), find the value of each individual resistance \( R \).
Given: \( R_s = 3 R = 45\,\Omega \)
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Detailed Solution 27
Individual Resistance (\(R\)):
\( 3 R = 45 \implies R = \frac{45}{3} = 15\,\Omega \)
\( 3 R = 45 \implies R = \frac{45}{3} = 15\,\Omega \)
QUESTION 28Decorative Lights FaultCBSE Conceptual
In a series decorative lighting string of 50 bulbs, one bulb fuses. What happens to the remaining 49 bulbs? Explain why series wiring is unsuitable for domestic lighting.
Concept: Series circuit single point of failure.
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Detailed Solution 28
Effect: All remaining 49 bulbs stop glowing because the series circuit becomes open (broken).
Domestic Unsuitability: In series, if one appliance fails, all turn off. Also, voltage gets divided among appliances.
Domestic Unsuitability: In series, if one appliance fails, all turn off. Also, voltage gets divided among appliances.
QUESTION 29Series Formula ProofCBSE Derivation
Derive the expression for the equivalent resistance (\(R_s\)) of three resistors \( R_1, R_2, R_3 \) connected in series across a voltage source \( V \).
Derivation: \( V = V_1 + V_2 + V_3 \), Current \(I\) constant.
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Detailed Solution 29
Derivation:
Total Potential Difference: \( V = V_1 + V_2 + V_3 \)
By Ohm's Law: \( V_1 = I R_1, V_2 = I R_2, V_3 = I R_3 \)
\( I R_s = I R_1 + I R_2 + I R_3 \)
\( R_s = R_1 + R_2 + R_3 \)
Total Potential Difference: \( V = V_1 + V_2 + V_3 \)
By Ohm's Law: \( V_1 = I R_1, V_2 = I R_2, V_3 = I R_3 \)
\( I R_s = I R_1 + I R_2 + I R_3 \)
\( R_s = R_1 + R_2 + R_3 \)
QUESTION 30Parallel CombinationCBSE Board Core
Two resistors of \( 6\,\Omega \) and \( 12\,\Omega \) are connected in parallel across a 12 V battery. Calculate:
(a) Equivalent parallel resistance (\(R_p\)).
(b) Total current drawn from battery.
(a) Equivalent parallel resistance (\(R_p\)).
(b) Total current drawn from battery.
Given: \( R_1 = 6\,\Omega, R_2 = 12\,\Omega, V = 12\text{ V} \)
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Detailed Solution 30
(a) Equivalent Resistance (\(R_p\)):
\( R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4.0\,\Omega \)
(b) Total Current (\(I\)):
\( I = \frac{V}{R_p} = \frac{12\text{ V}}{4\,\Omega} = 3.0\text{ Amperes} \)
\( R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4.0\,\Omega \)
(b) Total Current (\(I\)):
\( I = \frac{V}{R_p} = \frac{12\text{ V}}{4\,\Omega} = 3.0\text{ Amperes} \)
QUESTION 31Parallel Branch CurrentsCBSE NCERT Textbook
Three resistors \( 5\,\Omega, 10\,\Omega, \) and \( 30\,\Omega \) are connected in parallel across a 12 V battery. Find branch currents \( I_1, I_2, I_3 \) and total circuit resistance \( R_p \).
Given: \( R_1=5\,\Omega, R_2=10\,\Omega, R_3=30\,\Omega, V=12\text{ V} \)
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Detailed Solution 31
Branch Currents:
\( I_1 = 12/5 = 2.4\text{ A} \)
\( I_2 = 12/10 = 1.2\text{ A} \)
\( I_3 = 12/30 = 0.4\text{ A} \)
Total Current = 4.0 A
Equivalent Resistance (\(R_p\)):
\( R_p = \frac{12\text{ V}}{4.0\text{ A}} = 3.0\,\Omega \)
\( I_1 = 12/5 = 2.4\text{ A} \)
\( I_2 = 12/10 = 1.2\text{ A} \)
\( I_3 = 12/30 = 0.4\text{ A} \)
Total Current = 4.0 A
Equivalent Resistance (\(R_p\)):
\( R_p = \frac{12\text{ V}}{4.0\text{ A}} = 3.0\,\Omega \)
QUESTION 32Current Division RuleCBSE Board Core
Total current entering a parallel junction of \( 3\,\Omega \) and \( 6\,\Omega \) is \( I = 9 \text{ A} \). Calculate current flowing through each resistor.
Given: \( I = 9\text{ A}, R_1 = 3\,\Omega, R_2 = 6\,\Omega \)
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Detailed Solution 32
Current Division:
\( I_1 = I \times \frac{R_2}{R_1 + R_2} = 9 \times \frac{6}{3+6} = 6.0\text{ A} \)
\( I_2 = 9 - 6 = 3.0\text{ A} \)
\( I_1 = I \times \frac{R_2}{R_1 + R_2} = 9 \times \frac{6}{3+6} = 6.0\text{ A} \)
\( I_2 = 9 - 6 = 3.0\text{ A} \)
QUESTION 33Domestic Wiring PhysicsCBSE Core Theory
State three advantages of connecting electrical appliances in parallel rather than in series in domestic circuits.
Concept: Domestic parallel circuit advantages.
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Detailed Solution 33
Advantages:
1. Equal Voltage: Each appliance gets full supply voltage (220V).
2. Independent Operation: Each device can be turned ON/OFF independently.
3. Lower Total Resistance: Overall circuit resistance decreases, providing adequate power.
1. Equal Voltage: Each appliance gets full supply voltage (220V).
2. Independent Operation: Each device can be turned ON/OFF independently.
3. Lower Total Resistance: Overall circuit resistance decreases, providing adequate power.
QUESTION 34Parallel Formula ProofCBSE Derivation
Derive the expression for equivalent resistance (\(R_p\)) of two resistors \( R_1 \) and \( R_2 \) connected in parallel across voltage \( V \).
Derivation: \( I = I_1 + I_2 \), Voltage \(V\) constant.
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Detailed Solution 34
Derivation:
Total Current: \( I = I_1 + I_2 \)
By Ohm's Law: \( I_1 = \frac{V}{R_1}, I_2 = \frac{V}{R_2} \)
\( \frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} \implies \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \)
Total Current: \( I = I_1 + I_2 \)
By Ohm's Law: \( I_1 = \frac{V}{R_1}, I_2 = \frac{V}{R_2} \)
\( \frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} \implies \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \)
QUESTION 35Mixed Circuit NetworkCBSE High Yield
Two resistors of \( 8\,\Omega \) each are connected in parallel. This combination is connected in series with a \( 4\,\Omega \) resistor across a 12 V battery. Calculate total circuit resistance and current.
Given: \( R_p = 8 \parallel 8 = 4\,\Omega \), \( R_{series} = 4\,\Omega, V = 12\text{ V} \)
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Detailed Solution 35
Parallel Branch: \( R_{par} = \frac{8 \times 8}{8+8} = 4\,\Omega \)
Total Resistance: \( R_{tot} = 4 + 4 = 8.0\,\Omega \)
Total Current: \( I = \frac{12\text{ V}}{8\,\Omega} = 1.5\text{ A} \)
Total Resistance: \( R_{tot} = 4 + 4 = 8.0\,\Omega \)
Total Current: \( I = \frac{12\text{ V}}{8\,\Omega} = 1.5\text{ A} \)
QUESTION 36Ladder Circuit NetworkCBSE HOTS
Three resistors of \( 6\,\Omega \) each are to be combined. How would you connect them to get an equivalent resistance of: (a) \( 9\,\Omega \), (b) \( 4\,\Omega \)?
Given: Three \( 6\,\Omega \) resistors. Target resistances: 9 ohm and 4 ohm.
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Detailed Solution 36
(a) For 9 \(\Omega\): Connect two in parallel (\(6 \parallel 6 = 3\,\Omega\)) and third in series (\(3 + 6 = 9\,\Omega\)).
(b) For 4 \(\Omega\): Connect two in series (\(6 + 6 = 12\,\Omega\)) and third in parallel (\(12 \parallel 6 = \frac{72}{18} = 4\,\Omega\)).
(b) For 4 \(\Omega\): Connect two in series (\(6 + 6 = 12\,\Omega\)) and third in parallel (\(12 \parallel 6 = \frac{72}{18} = 4\,\Omega\)).
QUESTION 37Joule's Heating LawCBSE Core Theory
State Joule's Law of Heating. Write the mathematical formula for heat produced (\(H\)) in terms of current (\(I\)), resistance (\(R\)), and time (\(t\)).
Formula: \( H = I^2 R t \)
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Detailed Solution 37
Statement: Heat produced in a resistor is directly proportional to:
1. Square of current (\(I^2\))
2. Resistance (\(R\))
3. Time (\(t\))
Formula: \( H = I^2 R t = V I t = \frac{V^2}{R} t \)
1. Square of current (\(I^2\))
2. Resistance (\(R\))
3. Time (\(t\))
Formula: \( H = I^2 R t = V I t = \frac{V^2}{R} t \)
QUESTION 38Joule Heating NumericalCBSE NCERT Textbook
\( 100 \text{ J} \) of heat is produced each second in a \( 4 \,\Omega \) resistor. Find potential difference across the resistor.
Given: \( H/t = 100\text{ W} \), \( R = 4\,\Omega \)
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Detailed Solution 38
Potential Difference (\(V\)):
\( P = \frac{V^2}{R} \implies 100 = \frac{V^2}{4} \implies V^2 = 400 \implies V = 20\text{ Volts} \)
\( P = \frac{V^2}{R} \implies 100 = \frac{V^2}{4} \implies V^2 = 400 \implies V = 20\text{ Volts} \)
QUESTION 39Heater Cord PhysicsCBSE Conceptual
Why does the cord of an electric heater not glow while the heating element does?
Concept: Resistance difference between copper cord and Nichrome element.
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Detailed Solution 39
Explanation: Copper cord has very low resistance (\(R\)), so heat produced (\(I^2 R\)) is negligible. Nichrome heating element has high resistance, producing large heat energy and glowing red-hot.
QUESTION 40Electric Iron NumericalCBSE Board Core
An electric iron consumes energy at a rate of 840 W when heating is at maximum rate and 360 W when at minimum rate. The voltage is 220 V. Calculate current and resistance in each case.
Given: \( P_{max} = 840\text{ W}, P_{min} = 360\text{ W}, V = 220\text{ V} \)
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Detailed Solution 40
Max Rate: \( I = 840/220 = 3.82\text{ A} \), \( R = 220/3.82 = 57.6\,\Omega \)
Min Rate: \( I = 360/220 = 1.64\text{ A} \), \( R = 220/1.64 = 134.15\,\Omega \)
Min Rate: \( I = 360/220 = 1.64\text{ A} \), \( R = 220/1.64 = 134.15\,\Omega \)
QUESTION 41Fuse Wire RatingCBSE Safety Physics
An electric iron of rating 1 kW is operated in a 220 V circuit. Which fuse rating (1A, 2A, 3A, 5A) must be used? Explain.
Given: \( P = 1000\text{ W}, V = 220\text{ V} \)
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Detailed Solution 41
Operating Current: \( I = P/V = 1000/220 = 4.54\text{ A} \)
Fuse Selection: Use a 5 A fuse because it is slightly greater than 4.54 A.
Fuse Selection: Use a 5 A fuse because it is slightly greater than 4.54 A.
QUESTION 42Tungsten Filament PhysicsCBSE Conceptual
Why is tungsten used almost exclusively for filament of electric lamps? Why are lamps filled with inactive nitrogen/argon gas?
Concept: Melting point and gas filling in incandescent bulbs.
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Detailed Solution 42
Tungsten: Extremely high melting point (3380 °C) so it does not melt at white-hot temperatures.
Inactive Gas: Prevents oxidation and evaporation of tungsten filament.
Inactive Gas: Prevents oxidation and evaporation of tungsten filament.
QUESTION 43Series vs Parallel HeatingCBSE HOTS
Two identical resistance wires are connected first in series and then in parallel across the same voltage. Find the ratio of heat produced in series to parallel (\(H_s / H_p\)).
Given: Two identical wires \(R\). Voltage \(V\) same.
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Detailed Solution 43
Heat Ratio:
\( H_s = \frac{V^2}{2R} t \), \( H_p = \frac{V^2}{R/2} t = 2 \frac{V^2}{R} t \)
\( \frac{H_s}{H_p} = \frac{1/2}{2} = \frac{1}{4} \implies 1 : 4 \)
\( H_s = \frac{V^2}{2R} t \), \( H_p = \frac{V^2}{R/2} t = 2 \frac{V^2}{R} t \)
\( \frac{H_s}{H_p} = \frac{1/2}{2} = \frac{1}{4} \implies 1 : 4 \)
QUESTION 44Electric Power ConceptCBSE Board Core
Define electric power. State its SI unit. Write three different mathematical formulas to calculate electric power.
Formulas: \( P = VI = I^2 R = V^2 / R \)
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Detailed Solution 44
Definition: Rate at which electrical energy is consumed in a circuit.
SI Unit: Watt (W) = 1 Joule/second.
Formulas: \( P = VI \), \( P = I^2 R \), \( P = \frac{V^2}{R} \)
SI Unit: Watt (W) = 1 Joule/second.
Formulas: \( P = VI \), \( P = I^2 R \), \( P = \frac{V^2}{R} \)
QUESTION 45Appliance Power RatingCBSE Board Core
An electric lamp is rated 220 V, 100 W. Calculate:
(a) Its resistance.
(b) Current drawn when operated at 110 V.
(a) Its resistance.
(b) Current drawn when operated at 110 V.
Given: Rated \( V = 220\text{ V}, P = 100\text{ W} \). Operating \( V' = 110\text{ V} \)
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Detailed Solution 45
(a) Resistance (\(R\)):
\( R = \frac{V^2}{P} = \frac{220^2}{100} = 484\,\Omega \)
(b) Current at 110V:
\( I = \frac{V'}{R} = \frac{110}{484} = 0.227\text{ A} \) (Power at 110V = 25 W)
\( R = \frac{V^2}{P} = \frac{220^2}{100} = 484\,\Omega \)
(b) Current at 110V:
\( I = \frac{V'}{R} = \frac{110}{484} = 0.227\text{ A} \) (Power at 110V = 25 W)
QUESTION 46Bulb Brightness ComparisonCBSE HOTS
Two bulbs rated \( 60 \text{ W}, 220 \text{ V} \) and \( 100 \text{ W}, 220 \text{ V} \) are connected: (a) in parallel, (b) in series across 220 V. State which bulb glows brighter in each case.
Given: Bulb 1 (60W), Bulb 2 (100W). Voltage = 220V.
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Detailed Solution 46
(a) In Parallel: The 100W bulb glows brighter (higher power \(P = V^2/R\)).
(b) In Series: The 60W bulb glows brighter (since \(R_{60W} > R_{100W}\) and \(P = I^2 R\)).
(b) In Series: The 60W bulb glows brighter (since \(R_{60W} > R_{100W}\) and \(P = I^2 R\)).
QUESTION 471 kWh ConversionCBSE Board Standard
Define 1 kWh (commercial unit of electrical energy) and convert 1 kWh into SI unit Joules (\(\text{J}\)).
Concept: Commercial unit conversion.
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Detailed Solution 47
Definition: Energy consumed by a 1 kW appliance running for 1 hour.
Conversion:
\( 1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ Joules} \)
Conversion:
\( 1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ Joules} \)
QUESTION 48Electricity Bill 1CBSE Board Core
An electric refrigerator of 400 W operates 8 h/day and two fans of 80 W each operate 12 h/day. Calculate total energy consumed in 30 days and total cost at ₹ 4.00 per kWh.
Given: Fridge (400W, 8h), 2 Fans (160W, 12h), 30 days, ₹4/unit.
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Detailed Solution 48
Daily Energy:
Fridge: \( 0.4 \times 8 = 3.2\text{ kWh} \)
Fans: \( 2 \times 0.08 \times 12 = 1.92\text{ kWh} \)
Daily Total = 5.12 kWh
30-Day Cost:
Total Energy = \( 5.12 \times 30 = 153.6\text{ kWh} \)
Cost = \( 153.6 \times 4.00 = \text{₹ } 614.40 \)
Fridge: \( 0.4 \times 8 = 3.2\text{ kWh} \)
Fans: \( 2 \times 0.08 \times 12 = 1.92\text{ kWh} \)
Daily Total = 5.12 kWh
30-Day Cost:
Total Energy = \( 5.12 \times 30 = 153.6\text{ kWh} \)
Cost = \( 153.6 \times 4.00 = \text{₹ } 614.40 \)
QUESTION 49Electricity Bill 2CBSE High Yield
A house uses 4 bulbs of 40 W for 5 h/day, a TV of 100 W for 6 h/day, and a heater of 1000 W for 2 h/day. Calculate electricity bill for April (30 days) at ₹ 6.50 per unit.
Given: Bulbs (160W, 5h), TV (100W, 6h), Heater (1000W, 2h), ₹6.50/unit.
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Detailed Solution 49
Daily Energy:
Bulbs: \( 4 \times 0.04 \times 5 = 0.8\text{ kWh} \)
TV: \( 0.1 \times 6 = 0.6\text{ kWh} \)
Heater: \( 1.0 \times 2 = 2.0\text{ kWh} \)
Daily Total = 3.4 kWh
April Bill:
Total Units = \( 3.4 \times 30 = 102\text{ kWh} \)
Cost = \( 102 \times 6.50 = \text{₹ } 663.00 \)
Bulbs: \( 4 \times 0.04 \times 5 = 0.8\text{ kWh} \)
TV: \( 0.1 \times 6 = 0.6\text{ kWh} \)
Heater: \( 1.0 \times 2 = 2.0\text{ kWh} \)
Daily Total = 3.4 kWh
April Bill:
Total Units = \( 3.4 \times 30 = 102\text{ kWh} \)
Cost = \( 102 \times 6.50 = \text{₹ } 663.00 \)
QUESTION 50Case-Study CompetencyCBSE Competency
A modern home operates a total load of 2200 W on a 220 V mains line protected by a main fuse.
(a) Calculate main current drawn.
(b) Select fuse rating (5A, 10A, 15A).
(c) Calculate monthly energy units (5 h/day, 30 days).
(a) Calculate main current drawn.
(b) Select fuse rating (5A, 10A, 15A).
(c) Calculate monthly energy units (5 h/day, 30 days).
Given: \( P_{total} = 2200\text{ W}, V = 220\text{ V} \), 5 h/day, 30 days.
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Detailed Solution 50
(a) Main Current: \( I = 2200/220 = 10.0\text{ A} \)
(b) Fuse Rating: Select a 15 A fuse (slightly greater than 10 A operating current).
(c) Monthly Energy: Daily = \( 2.2 \times 5 = 11\text{ kWh} \). Monthly = \( 11 \times 30 = 330\text{ Units} \)
(b) Fuse Rating: Select a 15 A fuse (slightly greater than 10 A operating current).
(c) Monthly Energy: Daily = \( 2.2 \times 5 = 11\text{ kWh} \). Monthly = \( 11 \times 30 = 330\text{ Units} \)