Vardaan Learning Institute - Blackboard Classroom

QUESTION 01 Drift Velocity & Relaxation Time CBSE Board

A uniform copper conductor of cross-sectional area \( A = 1.0 \times 10^{-7} \text{ m}^2 \) carries a steady current of \( I = 1.5 \text{ A} \). Assuming electron density is \( n = 8.5 \times 10^{28} \text{ m}^{-3} \), calculate:
(a) Drift velocity (\(v_d\)) of free electrons.
(b) Current density (\(J\)) in the conductor.
(c) Relaxation time (\(\tau\)) if resistivity \(\rho = 1.7 \times 10^{-8} \,\Omega\cdot\text{m}\) and electron mass \(m = 9.1 \times 10^{-31} \text{ kg}\).

Given: \( I = 1.5\text{ A} \), \( A = 10^{-7}\text{ m}^2 \), \( n = 8.5 \times 10^{28}\text{ m}^{-3} \), \( e = 1.6 \times 10^{-19}\text{ C} \)

Drift Velocity & Relaxation Time Diagram

01 / 20

Detailed Solution 01

(a) Drift Velocity (\(v_d\)):
\( I = n e A v_d \implies v_d = \frac{I}{n e A} = \frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-7}} = 1.10 \text{ mm/s} \)

(b) Current Density (\(J\)):
\( J = \frac{I}{A} = \frac{1.5}{1.0 \times 10^{-7}} = 1.5 \times 10^7 \text{ A/m}^2 \)

(c) Relaxation Time (\(\tau\)):
\( \tau = \frac{m}{n e^2 \rho} = 2.47 \times 10^{-14} \text{ seconds} \)

QUESTION 02 Temperature Dependence CBSE / JEE Main

A platinum resistance thermometer has a resistance of \( R_0 = 5.00\,\Omega \) at \(0^\circ\text{C}\) and \( R_{100} = 5.23\,\Omega \) at \(100^\circ\text{C}\). When inserted into a hot bath, the resistance becomes \( R_t = 5.795\,\Omega \). Calculate:
(a) Temperature coefficient of resistance (\(\alpha\)) of platinum.
(b) Temperature (\(t\)) of the hot bath.

Given: \( R_0 = 5.00\,\Omega \), \( R_{100} = 5.23\,\Omega \), \( R_t = 5.795\,\Omega \)

02 / 20

Detailed Solution 02

(a) Temperature Coefficient (\(\alpha\)):
\( \alpha = \frac{R_{100} - R_0}{100 R_0} = \frac{0.23}{500} = 4.6 \times 10^{-4} \,^\circ\text{C}^{-1} \)

(b) Temperature of Hot Bath (\(t\)):
\( t = \frac{R_t - R_0}{\alpha R_0} = \frac{0.795}{0.23} \times 100 = 345.65^\circ\text{C} \)

QUESTION 03 Infinite Resistor Ladder JEE Mains Special

An infinite network of identical resistors, each of resistance \( R = 2.0\,\Omega \), is connected across terminals A and B as shown. Determine the equivalent resistance (\(R_{AB}\)) of this infinite ladder network.

Given: Each resistor \( R = 2.0\,\Omega \), Infinite repeating unit network.

03 / 20

Detailed Solution 03

Equivalent Resistance Method:
Let total equivalent resistance across A-B be \( R_{AB} = X \).
\( X = R + \frac{R X}{R + X} \implies X^2 - R X - R^2 = 0 \)
For \( R = 2.0\,\Omega \): \( X^2 - 2X - 4 = 0 \implies X = 1 + \sqrt{5} = 3.236 \,\Omega \)

QUESTION 04 Mixed Cell Grouping & Max Power CBSE / JEE Main

A battery consists of \( n = 4 \) parallel rows, where each row contains \( m = 6 \) identical cells connected in series. Each cell has EMF \( e = 2.0\text{ V} \) and internal resistance \( r = 0.5\,\Omega \).
(a) Determine total EMF (\(E_{eq}\)) and internal resistance (\(r_{eq}\)) of the battery.
(b) Find the external load resistance (\(R\)) for maximum power transfer and calculate maximum power (\(P_{max}\)).

Given: \( m = 6 \) (in series), \( n = 4 \) (parallel rows), \( e = 2.0\text{V} \), \( r = 0.5\,\Omega \).

04 / 20

Detailed Solution 04

(a) Battery Equivalent EMF & Resistance:
\( E_{eq} = m \cdot e = 6 \times 2.0 = 12\text{ V} \)
\( r_{eq} = \frac{m r}{n} = \frac{6 \times 0.5}{4} = 0.75\,\Omega \)

(b) Max Power Transfer:
\( R = r_{eq} = 0.75\,\Omega \)
\( I = \frac{12}{0.75 + 0.75} = 8\text{ A} \implies P_{max} = I^2 R = 64 \times 0.75 = 48\text{ Watts} \)

QUESTION 05 Kirchhoff's Rules (KVL / KCL) CBSE Board Core

In the multi-loop circuit shown, determine branch currents \(I_1\), \(I_2\), and \(I_3\) using Kirchhoff's Voltage Law (KVL).
(Given: Battery EMFs \(E_1 = 10\text{ V}\), \(E_2 = 4\text{ V}\); Resistors \(R_1 = 2\,\Omega\), \(R_2 = 4\,\Omega\), central resistor \(R_3 = 1\,\Omega\)).

Given: \( E_1 = 10\text{V}, E_2 = 4\text{V}, R_1 = 2\,\Omega, R_2 = 4\,\Omega, R_3 = 1\,\Omega \)

05 / 20

Detailed Solution 05

Applying KCL & KVL:
\( I_3 = I_1 + I_2 \)
Loop 1: \( 10 - 2 I_1 - 1(I_1 + I_2) = 0 \implies 3 I_1 + I_2 = 10 \)
Loop 2: \( 4 - 4 I_2 - 1(I_1 + I_2) = 0 \implies I_1 + 5 I_2 = 4 \)
Solving: \( I_2 = 0.143\text{ A} \), \( I_1 = 3.285\text{ A} \), \( I_3 = 3.428\text{ A} \)

QUESTION 06 Wheatstone Bridge Balance JEE Mains / CBSE

A Wheatstone bridge has four arms with resistances \( P = 10\,\Omega \), \( Q = 15\,\Omega \), \( R = 20\,\Omega \), and \( S = 30\,\Omega \). A galvanometer \( G = 50\,\Omega \) is connected across B-D and a battery of \( E = 6.0\text{ V} \) is across A-C.
(a) Prove whether the bridge is balanced.
(b) Find the total equivalent resistance (\(R_{AC}\)) and total current drawn from the battery.

Given: \( P=10\,\Omega, Q=15\,\Omega, R=20\,\Omega, S=30\,\Omega, E=6\text{V} \).

06 / 20

Detailed Solution 06

(a) Balance Condition:
\( \frac{P}{Q} = \frac{10}{15} = \frac{2}{3} \) and \( \frac{R}{S} = \frac{20}{30} = \frac{2}{3} \). Since \( \frac{P}{Q} = \frac{R}{S} \), the bridge is balanced (\(I_G = 0\)).

(b) Equivalent Resistance:
\( R_{AC} = \frac{(10+15)(20+30)}{(10+15)+(20+30)} = \frac{25 \times 50}{75} = 16.67 \,\Omega \)
\( I = \frac{6.0}{16.67} = 0.36\text{ A} \)

QUESTION 07 Meter Bridge Experiment CBSE Board Practical

In a meter bridge experiment, null point is obtained at \( l_1 = 33.7\text{ cm} \) from end A when a standard resistance \( R = 12.0\,\Omega \) is in the left gap and unknown resistance \( S \) is in the right gap.
(a) Calculate the unknown resistance \(S\).
(b) If \(R\) and \(S\) are interchanged, calculate the new balance length (\(l_2\)).
(c) Why are thick copper strips used to connect components in a meter bridge?

Given: Left gap \( R = 12.0\,\Omega \), Balance length \( l_1 = 33.7\text{ cm} \).

07 / 20

Detailed Solution 07

(a) Unknown Resistance (\(S\)):
\( \frac{12.0}{S} = \frac{33.7}{100 - 33.7} \implies S = 12.0 \times \frac{66.3}{33.7} = 23.61 \,\Omega \)

(b) Interchanging \(R\) and \(S\):
New balance length \( l_2 = 100 - 33.7 = 66.3\text{ cm} \)

(c) Copper Strips:
Thick copper strips have very large area and negligible resistance, reducing end resistance errors.

QUESTION 08 Potentiometer - EMF Comparison CBSE Board Classic

A potentiometer wire \( AB \) of length \( L = 100\text{ cm} \) has a resistance \( R_w = 10\,\Omega \). It is connected in series with driver cell \( E_0 = 4.0\text{ V} \) and series resistor \( R_s = 30\,\Omega \).
(a) Calculate potential gradient (\(k\)) along the wire.
(b) Two cells \(E_1\) and \(E_2\) produce balance lengths \(l_1 = 60\text{ cm}\) and \(l_2 = 45\text{ cm}\) respectively. Determine the ratio \(E_1 / E_2\) and exact value of \(E_1\).

Given: \( L = 100\text{cm}, R_w = 10\,\Omega, E_0 = 4\text{V}, R_s = 30\,\Omega, l_1 = 60\text{cm}, l_2 = 45\text{cm} \).

08 / 20

Detailed Solution 08

(a) Potential Gradient (\(k\)):
\( I_0 = \frac{4.0}{10 + 30} = 0.1\text{ A} \implies V_{AB} = 0.1 \times 10 = 1.0\text{ V} \)
\( k = \frac{1.0\text{ V}}{100\text{ cm}} = 0.01 \text{ V/cm} = 1.0 \text{ V/m} \)

(b) Ratio & Value of \(E_1\):
\( \frac{E_1}{E_2} = \frac{60}{45} = \frac{4}{3} = 1.33 \)
\( E_1 = k \cdot l_1 = 0.01 \times 60 = 0.60 \text{ Volts} \)

QUESTION 09 Potentiometer - Internal Resistance CBSE / JEE Main

A cell of EMF \( E = 1.5\text{ V} \) gives a balance length of \( l_1 = 75.0\text{ cm} \) on a potentiometer wire when the secondary circuit key is open. When a resistance box \( R = 9.5\,\Omega \) is shunted across the cell, the balance length reduces to \( l_2 = 60.0\text{ cm} \).
(a) Calculate internal resistance (\(r\)) of the cell.
(b) Explain why a potentiometer is preferred over a voltmeter for measuring exact EMF of a cell.

Given: \( l_1 = 75.0\text{cm} \) (open circuit), \( l_2 = 60.0\text{cm} \) (closed circuit), \( R = 9.5\,\Omega \).

Potentiometer Internal Resistance Diagram

09 / 20

Detailed Solution 09

(a) Internal Resistance (\(r\)):
\( r = R \left( \frac{l_1}{l_2} - 1 \right) = 9.5 \left( \frac{75.0}{60.0} - 1 \right) = 9.5 \times 0.25 = 2.375 \,\Omega \)

(b) Advantage of Potentiometer:
Potentiometer draws zero current at balance point (null deflection), measuring true EMF \(E\). A voltmeter draws current, measuring terminal voltage \(V < E\).

QUESTION 10 Frustum Conductor Analysis JEE Mains Advanced

A conductor of length \(L\) and non-uniform cross-section is in the shape of a truncated cone (frustum) with radii \(a\) and \(b\) (\(b > a\)) at its ends. It carries a steady current \(I\).
(a) Derive expression for total resistance (\(R\)) of the frustum in terms of resistivity \(\rho\), length \(L\), and radii \(a, b\).
(b) Determine how Electric Field \(E(x)\) and Drift Velocity \(v_d(x)\) vary along length \(x\) from end \(a\) to \(b\).

Given: Frustum conductor, Radii \(a\) and \(b\), Length \(L\), Resistivity \(\rho\).

10 / 20

Detailed Solution 10

(a) Frustum Resistance Integration:
Radius: \( r(x) = a + \frac{b - a}{L} x \)
\( dR = \frac{\rho dx}{\pi [r(x)]^2} \implies R = \int_0^L dR = \frac{\rho L}{\pi a b} \)

(b) Field & Velocity Variation:
\( E(x) = \frac{\rho I}{\pi r(x)^2} \propto \frac{1}{r(x)^2} \)
\( v_d(x) = \frac{I}{n e \pi r(x)^2} \propto \frac{1}{r(x)^2} \)

QUESTION 11 Cube Resistor Network JEE Mains / Advanced

Twelve identical resistors, each of resistance \(R = 6.0\,\Omega\), form a 3D skeletal cube. A potential difference \(V = 10\text{ V}\) is applied across the body diagonal terminals A and G.
(a) Calculate total equivalent resistance (\(R_{eq}\)) across body diagonal A-G.
(b) Calculate total current drawn from the voltage source.

Given: 12 cube resistors \(R = 6.0\,\Omega\), Body diagonal terminals A-G, \(V = 10\text{V}\).

11 / 20

Detailed Solution 11

(a) Symmetry Analysis:
Current entering A (\(3I\)) splits equally into 3 branches at A (\(I\) each), then into 6 edges (\(I/2\) each), and recombines at G (\(3I\)).
Total Potential Drop: \( V = I R + \frac{I}{2} R + I R = \frac{5}{2} I R \)
Since total current is \(I_{total} = 3I\):
\( R_{eq} = \frac{V}{I_{total}} = \frac{\frac{5}{2} I R}{3 I} = \frac{5}{6} R \)
For \(R = 6.0\,\Omega\): \( R_{eq} = \frac{5}{6} \times 6 = 5.0\,\Omega \)

(b) Total Current:
\( I_{total} = \frac{V}{R_{eq}} = \frac{10}{5.0} = 2.0\text{ Amperes} \)

QUESTION 12 Delta-Wye (Δ-Y) Conversion JEE Advanced Special

In an unbalanced bridge circuit where central branch resistors form a Delta loop (\(R_A=3\Omega, R_B=6\Omega, R_C=3\Omega\)), apply Delta-Wye (\(\Delta\)-Y) transformation.
(a) Calculate equivalent Wye resistors (\(R_1, R_2, R_3\)).
(b) Determine total equivalent circuit resistance.

Given: Delta loop resistors \(R_A=3\Omega, R_B=6\Omega, R_C=3\Omega\).

12 / 20

Detailed Solution 12

Delta-Wye Formula:
Sum of Delta resistors: \( S_{sum} = R_A + R_B + R_C = 3 + 6 + 3 = 12\,\Omega \)
\( R_1 = \frac{R_A R_B}{S_{sum}} = \frac{3 \times 6}{12} = 1.5\,\Omega \)
\( R_2 = \frac{R_A R_C}{S_{sum}} = \frac{3 \times 3}{12} = 0.75\,\Omega \)
\( R_3 = \frac{R_B R_C}{S_{sum}} = \frac{6 \times 3}{12} = 1.5\,\Omega \)

QUESTION 13 Symmetrical Lattice Network JEE Mains / Advanced

A symmetrical bridge lattice network consists of 5 resistors each of resistance \(R = 4.0\,\Omega\) connected across A and B.
(a) Identify equipotential nodes using mirror symmetry.
(b) Calculate equivalent resistance (\(R_{AB}\)).

Given: 5 identical resistors \(R = 4.0\,\Omega\), Mirror symmetry along perpendicular axis.

13 / 20

Detailed Solution 13

Mirror Symmetry Method:
Perpendicular axis through central resistor bisects the circuit into symmetric halves.
Top and bottom nodes are at equal potential, so zero current flows through central vertical resistor.
Removing central resistor gives two parallel branches of \(2R\) each:
\( R_{AB} = \frac{2R \times 2R}{2R + 2R} = R = 4.0\,\Omega \)

QUESTION 14 Real Battery Charging vs Discharging CBSE Board Core

A storage battery of EMF \(E = 12\text{ V}\) and internal resistance \(r = 0.5\,\Omega\) is being charged by a \(120\text{ V}\) DC supply using a series resistor \(R = 11.5\,\Omega\).
(a) Calculate charging current (\(I\)).
(b) Calculate terminal voltage (\(V_{term}\)) of the battery during charging.
(c) What is the purpose of the series resistor \(R\)?

Given: \( E = 12\text{V}, r = 0.5\,\Omega, V_{supply} = 120\text{V}, R = 11.5\,\Omega \).

14 / 20

Detailed Solution 14

(a) Charging Current (\(I\)):
Net effective EMF: \( E_{net} = V_{supply} - E = 120 - 12 = 108\text{ V} \)
Total resistance: \( R_{total} = R + r = 11.5 + 0.5 = 12.0\,\Omega \)
\( I = \frac{108}{12.0} = 9.0\text{ Amperes} \)

(b) Terminal Voltage during Charging:
\( V_{term} = E + I r = 12 + (9.0 \times 0.5) = 12 + 4.5 = 16.5\text{ Volts} \)

(c) Purpose of Series Resistor:
Limits high charging current to prevent damage to battery and supply.

QUESTION 15 RC Circuit Transient Analysis JEE Mains Special

A capacitor of capacitance \(C = 100\,\mu\text{F}\) is charged through a resistor \(R = 20\,\text{k}\Omega\) by a battery of \(V_0 = 12\text{ V}\).
(a) Calculate time constant (\(\tau = RC\)) of the circuit.
(b) Calculate charge \(q(t)\) on the capacitor at \(t = \tau\).
(c) Calculate maximum current \(I_{max}\) at \(t = 0\).

Given: \( C = 100\,\mu\text{F}, R = 20\,\text{k}\Omega, V_0 = 12\text{V} \).

15 / 20

Detailed Solution 15

(a) Time Constant (\(\tau\)):
\( \tau = R C = (20 \times 10^3 \,\Omega) \times (100 \times 10^{-6} \text{ F}) = 2.0\text{ seconds} \)

(b) Charge at \(t = \tau\):
Max charge \( Q_0 = C V_0 = 100 \mu\text{F} \times 12\text{V} = 1.2 \text{ mC} \)
\( q(\tau) = Q_0 (1 - e^{-1}) = 1.2 \times (1 - 0.368) = 0.758 \text{ mC} \) (63.2% of max charge)

(c) Maximum Current:
\( I_{max} = \frac{V_0}{R} = \frac{12}{20 \times 10^3} = 0.6 \text{ mA} \)

QUESTION 16 Non-Uniform Resistivity Integration JEE Advanced Special

A cylindrical conductor of length \(L = 2.0\text{ m}\) and cross-sectional area \(A = 4.0 \times 10^{-6} \text{ m}^2\) has non-uniform resistivity varying along its length as \(\rho(x) = \rho_0 (1 + \alpha x)\), where \(\rho_0 = 1.0 \times 10^{-6} \,\Omega\cdot\text{m}\) and \(\alpha = 0.5\text{ m}^{-1}\).
(a) Derive and calculate total resistance (\(R\)) of the conductor.
(b) Find potential difference across first half (\(x = 0\) to \(1.0\text{m}\)) when carrying current \(I = 2\text{A}\).

Given: \( L = 2.0\text{m}, A = 4 \times 10^{-6}\text{m}^2, \rho_0 = 10^{-6}\,\Omega\cdot\text{m}, \alpha = 0.5\text{m}^{-1} \).

Non-Uniform Resistivity Conductor Diagram

16 / 20

Detailed Solution 16

(a) Total Resistance Integration:
\( R = \int_0^L \frac{\rho_0 (1 + \alpha x) dx}{A} = \frac{\rho_0 L}{A} \left( 1 + \frac{\alpha L}{2} \right) \)
\( R = \frac{10^{-6} \times 2.0}{4.0 \times 10^{-6}} \left( 1 + \frac{0.5 \times 2.0}{2} \right) = 0.5 \times 1.5 = 0.75 \,\Omega \)

(b) Voltage drop across first half:
\( R_{half} = \frac{10^{-6} \times 1.0}{4.0 \times 10^{-6}} \left( 1 + \frac{0.5 \times 1.0}{2} \right) = 0.3125 \,\Omega \)
\( V_{half} = 2.0 \times 0.3125 = 0.625\text{ Volts} \)

QUESTION 17 Carey Foster Bridge CBSE Practical / JEE

In a Carey Foster bridge, two nearly equal resistances \(X\) and \(Y\) give balance length \(l_1 = 48.5\text{ cm}\). When \(X\) and \(Y\) are interchanged, balance length becomes \(l_2 = 51.5\text{ cm}\). Resistance per unit length of bridge wire is \(\rho = 0.05\,\Omega/\text{cm}\).
(a) Calculate difference in resistance \(X - Y\).
(b) State the advantage of Carey Foster bridge over standard meter bridge.

Given: \( l_1 = 48.5\text{cm}, l_2 = 51.5\text{cm}, \rho = 0.05\,\Omega/\text{cm} \).

17 / 20

Detailed Solution 17

(a) Resistance Difference Calculation:
Formula: \( X - Y = \rho (l_2 - l_1) = 0.05 \times (51.5 - 48.5) = 0.05 \times 3.0 = 0.15 \,\Omega \)

(b) Advantage:
Eliminating end strip resistance errors completely, enabling precise measurement of small resistance differences.

QUESTION 18 Battery Terminal Voltage (V - I Graph) CBSE / JEE Main

The terminal voltage \(V\) of a cell as a function of current \(I\) drawn from it is plotted as a straight line. The line intersects the V-axis at \(V = 2.1\text{ V}\) and I-axis at \(I = 4.2\text{ A}\).
(a) Determine EMF (\(E\)) of the cell.
(b) Calculate internal resistance (\(r\)) of the cell.
(c) Calculate maximum short-circuit current.

Given: V-axis intercept = 2.1V, I-axis intercept = 4.2A.

18 / 20

Detailed Solution 18

(a) Cell EMF (\(E\)):
When \(I = 0\), \(V = E = 2.1\text{ Volts}\).

(b) Internal Resistance (\(r\)):
\( V = E - I r \implies r = \frac{2.1}{4.2} = 0.5 \,\Omega \)

(c) Short-Circuit Current:
\( I_{sc} = \frac{E}{r} = \frac{2.1}{0.5} = 4.2\text{ Amperes} \)

QUESTION 19 Potentiometer Sensitivity Tuning JEE Mains Special

A potentiometer has a 10-meter long wire of resistance \(R_w = 20\,\Omega\) connected to a \(2.0\text{V}\) battery.
(a) How can the sensitivity of this potentiometer be increased?
(b) Calculate new potential gradient (\(k'\)) if a series resistor \(R_s = 80\,\Omega\) is added to the primary circuit.

Given: \( L = 10\text{m}, R_w = 20\,\Omega, E_0 = 2.0\text{V}, R_s = 80\,\Omega \).

19 / 20

Detailed Solution 19

(a) Sensitivity Tuning:
Sensitivity increases by reducing potential gradient \(k\), achieved by increasing wire length \(L\) or adding series resistance \(R_s\).

(b) New Potential Gradient (\(k'\)):
\( I_0 = \frac{2.0}{20 + 80} = 0.02\text{ A} \implies V_{AB} = 0.02 \times 20 = 0.40\text{ V} \)
\( k' = \frac{0.40\text{ V}}{10\text{ m}} = 0.04 \text{ V/m} = 0.4 \text{ mV/cm} \)

QUESTION 20 Carbon Resistor Color Coding CBSE Board Core

A carbon composition resistor has four color bands in sequence: Yellow, Violet, Brown, and Gold.
(a) Determine its resistance value and percentage tolerance.
(b) Calculate the range within which the actual resistance lies.

Given: Color Code: Band 1 = Yellow (4), Band 2 = Violet (7), Band 3 = Brown (10¹), Band 4 = Gold (±5%).

20 / 20

Detailed Solution 20

(a) Color Code Resistance:
Resistance = \(47 \times 10^1 \,\Omega \pm 5\% = 470 \,\Omega \pm 5\%\)

(b) Tolerance Range:
5% of 470 = \(23.5 \,\Omega\)
Range: \(446.5 \,\Omega \text{ to } 493.5 \,\Omega\)