Vardaan Learning Institute

SI Unit: Ampere ($\text{A}$).
Definition of 1 Ampere: Mathematically, $I = Q/t$. Thus, $1\text{ A} = 1\text{ C} / 1\text{ s}$. Theoretically, when one coulomb of charge flows through a cross-section of a conductor in one second, the current is said to be one ampere.

Given: Total charge $Q = 1\text{ C}$, elementary charge $e = 1.6 \times 10^{-19}\text{ C}$.
Number of electrons, $n = \frac{Q}{e} = \frac{1}{1.6 \times 10^{-19}}$
$n = 0.625 \times 10^{19} = 6.25 \times 10^{18}\text{ electrons}$.

Given: Current $I = 0.5\text{ A}$, Time $t = 10\text{ minutes} = 10 \times 60\text{ s} = 600\text{ s}$.
Charge $Q = 0.5\text{ A} \times 600\text{ s} = 300\text{ Coulombs (C)}$.

Direction of Electrons: Since electrons are negatively charged, they actually flow from the negative terminal to the positive terminal of the battery. Thus, the conventional current direction is opposite to the actual flow of electrons.

SI Unit: Volt ($\text{V}$).
Definition of 1 Volt: $1\text{ Volt}$ is the potential difference when $1\text{ Joule}$ of work is done to move a charge of $1\text{ Coulomb}$ from one point to another ($1\text{ V} = 1\text{ J} / 1\text{ C}$).

Given: Charge $Q = 2\text{ C}$, Potential difference $V = 12\text{ V}$.
Work done $W = 12\text{ V} \times 2\text{ C} = 24\text{ Joules (J)}$.

How it works: The chemical action within the cell generates a potential difference across the terminals of the cell. This potential difference sets the electrons in motion and maintains the flow of electric current in the circuit.

Closed Circuit: A complete path where the key/switch is closed, allowing current to flow and appliances to work.
Open Circuit: A broken path where the key/switch is open (or a wire is broken), so no current flows and appliances stop working.

1. Straight line with a zigzag pattern: Resistor / Fixed Resistance.
2. Circle with 'A' inside it: Ammeter.
3. Combination of long and short parallel lines: Battery (combination of cells).
4. Circle with a cross (X) inside it: Electric bulb.

Voltmeter: Connected in parallel across the points where the potential difference is to be measured. It must have very high resistance so it draws negligible current from the main circuit, ensuring an accurate voltage reading.

Mathematical expression: $V \propto I \implies V = IR$, where $R$ is a constant called resistance.
Necessary conditions: The physical conditions like temperature, mechanical strain, and pressure must remain constant.

Slope representation: The slope of a V-I graph ($\frac{\Delta V}{\Delta I}$) represents the Resistance ($R$) of the conductor. Steeper slope = higher resistance. (Note: If it were an I-V graph, the slope would be $1/R$).

Cause of resistance: As free electrons drift through the metallic conductor, they continuously collide with the vibrating positive ions (atoms) of the metal lattice. These collisions obstruct their motion, causing electrical resistance.

$R = \frac{V_1}{I_1} = \frac{60\text{ V}}{4\text{ A}} = 15\text{ }\Omega$.
Now, use the new potential difference to find the new current ($I_2$):
$I_2 = \frac{V_2}{R} = \frac{120\text{ V}}{15\text{ }\Omega} = 8\text{ A}$.
The heater will draw $8\text{ A}$ of current.

1. Length of the conductor ($l$): $R \propto l$ (Directly proportional).
2. Area of cross-section ($A$): $R \propto \frac{1}{A}$ (Inversely proportional).
3. Nature of the material (Resistivity, $\rho$) and temperature.

Mathematical expression: $R = \rho \frac{l}{A}$

SI Unit: Ohm-meter ($\Omega\cdot\text{m}$).
Dependence: Resistivity is a property of the material. It does not depend on the length or thickness (area of cross-section) of the wire. It only changes with temperature and the nature of the material.

Initial resistance $R = \rho \frac{l}{A}$.
New length $l' = 2l$. Since volume $V = A \times l$ is constant, the new area $A' = A/2$.
New resistance $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{(A/2)} = 4 \left(\rho \frac{l}{A}\right) = 4R$.
The new resistance becomes four times the original resistance.

For the first wire: $R_1 = \rho \frac{l}{A} = 4\text{ }\Omega$.
For the second wire, length $l_2 = l/2$ and Area $A_2 = 2A$.
$R_2 = \rho \frac{l/2}{2A} = \frac{1}{4} \left(\rho \frac{l}{A}\right)$
$R_2 = \frac{1}{4} \times 4\text{ }\Omega = 1\text{ }\Omega$.
The resistance of the new wire is $1\text{ }\Omega$.

1. Higher Resistivity: Alloys generally have a higher resistivity than their constituent pure metals, allowing them to generate more heat ($H = I^2Rt$) for a given current.
2. Resistance to Oxidation: Alloys do not oxidize (burn) readily even at very high temperatures (when red hot), giving them a longer lifespan in heating appliances.

Given: $l = 1\text{ m}$, $R = 26\text{ }\Omega$.
Diameter $d = 0.3\text{ mm} = 0.3 \times 10^{-3}\text{ m}$.
Radius $r = d/2 = 0.15 \times 10^{-3}\text{ m}$.
Area $A = \pi r^2 = 3.14 \times (0.15 \times 10^{-3})^2 = 3.14 \times 0.0225 \times 10^{-6} = 7.065 \times 10^{-8}\text{ m}^2$.
Resistivity $\rho = \frac{26 \times 7.065 \times 10^{-8}}{1} = 183.69 \times 10^{-8}\text{ }\Omega\cdot\text{m} = 1.84 \times 10^{-6}\text{ }\Omega\cdot\text{m}$.

Let total voltage be $V = V_1 + V_2 + V_3$.
By Ohm's law, $V = IR_s$, $V_1 = IR_1$, $V_2 = IR_2$, $V_3 = IR_3$.
Substitute into the voltage equation:
$IR_s = IR_1 + IR_2 + IR_3$
$IR_s = I(R_1 + R_2 + R_3)$
Cancelling $I$ from both sides: $R_s = R_1 + R_2 + R_3$.

Let total current be $I = I_1 + I_2 + I_3$.
By Ohm's law, $I = V/R_p$, $I_1 = V/R_1$, $I_2 = V/R_2$, $I_3 = V/R_3$.
Substitute into the current equation:
$\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$
$\frac{V}{R_p} = V \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right)$
Cancelling $V$ from both sides: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.

1. If one appliance fails or is switched off, the others continue to work without interruption (because parallel branches are independent).
2. Each appliance gets the full source voltage ($220\text{ V}$ in India), allowing them to operate at their designed power ratings.
3. The total equivalent resistance of the circuit decreases, meaning the main supply can provide the high current needed for heavy appliances simultaneously.

1. Highest total resistance: Achieved by connecting all in series.
   $R_s = R_1 + R_2 + R_3 + R_4 = 4 + 8 + 12 + 24 = 48\text{ }\Omega$.
2. Lowest total resistance: Achieved by connecting all in parallel.
   $\frac{1}{R_p} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac{1}{2}\text{ }\Omega^{-1}$.
   $R_p = 2\text{ }\Omega$.

1. To get $9\text{ }\Omega$: Connect two $6\text{ }\Omega$ resistors in parallel, and the third in series with this combination.
   Parallel part: $R_p = \frac{6 \times 6}{6 + 6} = \frac{36}{12} = 3\text{ }\Omega$.
   Total $R = 3\text{ }\Omega + 6\text{ }\Omega \text{ (series)} = 9\text{ }\Omega$.
2. To get $4\text{ }\Omega$: Connect two $6\text{ }\Omega$ resistors in series, and the third in parallel with this combination.
   Series part: $R_s = 6 + 6 = 12\text{ }\Omega$.
   Parallel with third: $R_{total} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\text{ }\Omega$.

Upper branch: $R_1 = 10\text{ }\Omega$ and $R_2 = 20\text{ }\Omega$ are in series.
Equivalent resistance of upper branch, $R_s = 10 + 20 = 30\text{ }\Omega$.
Lower branch: $R_3 = 30\text{ }\Omega$.
Now, the upper branch ($30\text{ }\Omega$) and lower branch ($30\text{ }\Omega$) are in parallel between A and B.
$\frac{1}{R_p} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}\text{ }\Omega^{-1}$.
Equivalent resistance between A and B is $R_p = 15\text{ }\Omega$.

1. Total resistance ($R_p$):
   $\frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} = \frac{6 + 3 + 1}{30} = \frac{10}{30} = \frac{1}{3}\text{ }\Omega^{-1}$.
   $R_p = 3\text{ }\Omega$.
2. Total current ($I$):
   $I = \frac{V}{R_p} = \frac{12\text{ V}}{3\text{ }\Omega} = 4\text{ A}$.
3. Current through $5\text{ }\Omega$ resistor ($I_1$):
   In parallel, potential difference across each resistor is $12\text{ V}$.
   $I_1 = \frac{V}{R_1} = \frac{12}{5} = 2.4\text{ A}$.

In Series: If one bulb fuses, the circuit is broken (an open circuit). Current stops flowing entirely, so the remaining bulb will stop glowing completely.
In Parallel: If one bulb fuses, the branch it is in becomes open, but the other branch remains closed. The remaining bulb will continue to glow with the same brightness because the voltage across it remains the same.

Total resistance = $20\text{ }\Omega$.
Resistance of each semicircle = $20 / 2 = 10\text{ }\Omega$.
These two $10\text{ }\Omega$ resistors are in parallel.
$R_{eff} = \frac{10 \times 10}{10 + 10} = \frac{100}{20} = 5\text{ }\Omega$.
The effective resistance is $5\text{ }\Omega$.

In series: $R_s = R + R + ... \text{ (n times)} = nR$.
In parallel: $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + ... \text{ (n times)} = \frac{n}{R} \implies R_p = \frac{R}{n}$.
Ratio: $\frac{R_s}{R_p} = \frac{nR}{R/n} = n^2$.
The ratio is $n^2 : 1$.

1. Directly proportional to the square of the current ($I^2$) for a given resistance.
2. Directly proportional to the resistance ($R$) for a given current.
3. Directly proportional to the time ($t$) for which the current flows through the resistor.

Mathematical expression: $H = I^2Rt$.

Consider a current $I$ flowing through a resistor of resistance $R$. Let $V$ be the potential difference across it, and $t$ be the time.
Charge flowing in time $t$ is $Q = It$.
Work done ($W$) in moving charge $Q$ through potential difference $V$ is:
$W = V \times Q$
Substitute $Q = It$: $W = V \times I \times t$.
This work done is dissipated as heat ($H$) in the resistor. So, $H = VIt$.
According to Ohm's law, $V = IR$. Substituting this in the heat equation:
$H = (IR) \times I \times t \implies H = I^2Rt$.

1. Extremely high melting point ($3380^\circ\text{C}$): It can be heated to very high temperatures (becoming white-hot and emitting light) without melting.
2. High resistivity: It retains heat easily and allows the filament to reach light-emitting temperatures quickly without requiring a very long wire.

By filling the bulb with chemically inactive (inert) gases like argon or nitrogen, oxidation is prevented. This significantly prolongs the life of the filament.

Working Principle: It works on the heating effect of electric current. A fuse is a piece of wire of a material with an appropriate melting point (e.g., alloy of lead and tin). If current exceeds the safe limit, Joule heating ($H = I^2Rt$) melts the fuse wire, breaking the circuit.
Why in series? It must be in series with the live wire so that all the current entering the appliance passes through the fuse first. If it melts, it immediately cuts off power to the device.

Given: Resistance $R = 20\text{ }\Omega$, Current $I = 5\text{ A}$, Time $t = 30\text{ s}$.
$H = (5)^2 \times 20 \times 30$
$H = 25 \times 20 \times 30 = 500 \times 30 = 15000\text{ J}$.
The heat developed is $15000\text{ J}$ (or $15\text{ kJ}$).

Given: Heat $H = 100\text{ J}$, Resistance $R = 4\text{ }\Omega$, Time $t = 1\text{ s}$.
$100 = \frac{V^2}{4} \times 1$
$V^2 = 400$
$V = \sqrt{400} = 20\text{ V}$.
The potential difference across the resistor is $20\text{ V}$.

Given: Charge $Q = 96000\text{ C}$, Potential difference $V = 50\text{ V}$. (Time is extra information here if we use this formula directly).
$H = 50\text{ V} \times 96000\text{ C} = 4,800,000\text{ J} = 4.8 \times 10^6\text{ J}$.
The heat generated is $4.8 \times 10^6\text{ J}$.

Therefore, $H \propto R$. The heating element is made of an alloy (like nichrome) which has very high resistance, so it produces a massive amount of heat and glows red hot. The connecting cord is made of copper, which has very low resistance. Hence, it produces negligible heat and does not glow.

Since $V$ is constant, we use the formula $H = \frac{V^2}{R_{eq}}t$.
In series: $R_s = R + R = 2R$.
Heat in series, $H_s = \frac{V^2}{2R}t$.
In parallel: $R_p = \frac{R}{2}$.
Heat in parallel, $H_p = \frac{V^2}{R/2}t = \frac{2V^2}{R}t$.
Ratio: $\frac{H_s}{H_p} = \frac{\frac{V^2}{2R}t}{\frac{2V^2}{R}t} = \frac{1/2}{2} = \frac{1}{4}$.
The ratio is $1:4$.

SI Unit: Watt ($\text{W}$). One Watt is the power consumed by a device that carries $1\text{ A}$ of current when operated at a potential difference of $1\text{ V}$.

Work done (Energy) $W = VIt$. Power $P = W/t \implies P = \frac{VIt}{t} \implies \mathbf{P = VI}$.
Using Ohm's law ($V = IR$), substitute $V$: $P = (IR)I \implies \mathbf{P = I^2R}$.
Using Ohm's law ($I = V/R$), substitute $I$: $P = V(V/R) \implies \mathbf{P = V^2/R}$.
Usage: $P = I^2R$ is best used for series circuits because current ($I$) is constant. $P = V^2/R$ is best used for parallel circuits because voltage ($V$) is constant.

Relationship with Joules:
$1\text{ kWh} = 1\text{ kW} \times 1\text{ hour}$
$1\text{ kWh} = 1000\text{ W} \times 3600\text{ s}$
$1\text{ kWh} = 3.6 \times 10^6\text{ Watt-seconds} = \mathbf{3.6 \times 10^6\text{ Joules}}$.

From $P = \frac{V^2}{R} \implies R = \frac{V_{rating}^2}{P_{rating}}$.
$R = \frac{(220)^2}{100} = \frac{48400}{100} = 484\text{ }\Omega$.
Now, when operated at $V_{new} = 110\text{ V}$:
$P_{new} = \frac{V_{new}^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25\text{ W}$.
(Alternatively, since voltage is halved, power $P \propto V^2$ becomes $1/4$th of original: $100/4 = 25\text{ W}$).

Power $P = 400\text{ W} = 0.4\text{ kW}$.
Total time $t = 8\text{ hours/day} \times 30\text{ days} = 240\text{ hours}$.
Energy $E = P \times t = 0.4\text{ kW} \times 240\text{ h} = 96\text{ kWh}$.
Cost = Energy $\times$ Rate = $96\text{ kWh} \times \text{Rs } 3.00/\text{kWh} = \text{Rs } 288.00$.

Total maximum power allowed = $V \times I_{max} = 220\text{ V} \times 5\text{ A} = 1100\text{ W}$.
Power of one lamp = $10\text{ W}$.
Total power = $n \times (\text{Power of one lamp})$.
$1100\text{ W} = n \times 10\text{ W} \implies n = \frac{1100}{10} = 110$.
Therefore, 110 lamps can be connected in parallel.

First, find resistances: $R = V^2/P$.
Resistance of 100W bulb: $R_1 = (220)^2 / 100 = 484\text{ }\Omega$.
Resistance of 60W bulb: $R_2 = (220)^2 / 60 = 806.6\text{ }\Omega$.
So, $R_2 > R_1$ (lower wattage has higher resistance).
In series, the current $I$ is constant. Power consumed is $P = I^2R$.
Since $P \propto R$ in series, the bulb with higher resistance (the 60W bulb) will consume more power and glow brighter.

Case (i): $6\text{ V}$ battery in series with $1\text{ }\Omega$ and $2\text{ }\Omega$.
Total resistance $R_s = 1 + 2 = 3\text{ }\Omega$.
Current $I = V / R_s = 6 / 3 = 2\text{ A}$.
Power in $2\text{ }\Omega$ resistor: $P_1 = I^2R = (2)^2 \times 2 = 4 \times 2 = 8\text{ W}$.
Case (ii): $4\text{ V}$ battery in parallel with $12\text{ }\Omega$ and $2\text{ }\Omega$.
In parallel, the voltage across the $2\text{ }\Omega$ resistor is exactly $4\text{ V}$.
Power in $2\text{ }\Omega$ resistor: $P_2 = V^2 / R = (4)^2 / 2 = 16 / 2 = 8\text{ W}$.
Comparison: The power used by the $2\text{ }\Omega$ resistor is $8\text{ W}$ in both cases, so they are equal ($1:1$).

1. Mistake: Arnav has connected the Voltmeter in series and the Ammeter in parallel. They must be swapped.
2. Effect on current: A voltmeter has a very high resistance. When connected in series, it drastically increases the total resistance of the circuit, dropping the current to almost zero. An ammeter has low resistance; connecting it in parallel creates a short circuit across the resistor, further disrupting the measurement.
3. Correct circuit: The Ammeter must be placed in the main rectangular loop (in series with the battery and resistor), and the Voltmeter must be connected in a separate branch across (in parallel with) the resistor.

1. Total power: $P_{total} = P_1 + P_2 + P_3 = 2000\text{ W} + 1000\text{ W} + 1500\text{ W} = 4500\text{ W}$.
2. Total current: $I_{total} = \frac{P_{total}}{V} = \frac{4500\text{ W}}{220\text{ V}} = 20.45\text{ A}$.
3. Scientific explanation: The total current drawn ($20.45\text{ A}$) exceeds the safe limit of the $15\text{ A}$ fuse protecting the socket. The excessive current caused Joule heating, which melted the fuse wire, breaking the circuit. This phenomenon is called overloading.

1. Higher resistance: The slope of a V-I graph represents resistance ($R = V/I$). Since line A is steeper, it has a higher slope, meaning wire A has a higher resistance.
2. Thicker wire: Resistance is inversely proportional to cross-sectional area ($R \propto 1/A$). Since wire B has a lower resistance (gentler slope), it must have a larger cross-sectional area. Therefore, wire B is thicker.
3. Series combination: In series, the equivalent resistance ($R_s = R_A + R_B$) is greater than both individual resistances. The V-I graph for the series combination will have an even higher slope and will lie above line A (steeper than A).

These 5 parts are connected in parallel.
$\frac{1}{R'} = \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5}$
$\frac{1}{R'} = \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} = \frac{25}{R}$
Therefore, $R' = \frac{R}{25}$.
The ratio $\frac{R}{R'} = \frac{R}{R/25} = 25$.

Equivalent resistance of $n$ resistors in parallel is $R_p = \frac{r}{n} = \frac{176}{n}\text{ }\Omega$.
By Ohm's law, $R_p = \frac{V}{I}$.
$\frac{176}{n} = \frac{220\text{ V}}{5\text{ A}}$
$\frac{176}{n} = 44$
$n = \frac{176}{44} = 4$.
Therefore, 4 resistors are required.

However, if connected in series with other resistors, the total current $I$ in the circuit depends on the equivalent resistance. The heat dissipated by this specific resistor is $H = I^2Rt$. While increasing its resistance $R$ tends to increase heat, it also decreases the overall circuit current $I$ (which is squared). The actual effect depends on the relative magnitudes, but generally, in a purely series circuit with a constant voltage source, increasing one resistance will eventually decrease its own power dissipation after a certain peak (Maximum Power Transfer Theorem), but for small changes where $I$ is considered roughly constant relative to $R$, heat might seem to increase. The key is distinguishing between constant $V$ (parallel/single) and constant $I$ (series).

(i) Total resistance: Parallel part $R_p = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4\text{ }\Omega$.
Total resistance $R_{eq} = 4\text{ }\Omega\text{ (series part)} + 4\text{ }\Omega\text{ (parallel part)} = 8\text{ }\Omega$.
(ii) Voltage drop across $4\text{ }\Omega$ resistor: First find total current $I = V / R_{eq} = 12 / 8 = 1.5\text{ A}$.
Voltage drop $V_1 = I \times R_1 = 1.5\text{ A} \times 4\text{ }\Omega = 6\text{ V}$.
(iii) Heat by parallel combo: The parallel combination has equivalent resistance $R_p = 4\text{ }\Omega$ and current $I = 1.5\text{ A}$ flowing through it.
$H = I^2R_pt = (1.5)^2 \times 4 \times 10 = 2.25 \times 40 = 90\text{ Joules}$.

However, the bulb lights up almost instantly because turning on the switch establishes an electric field throughout the entire circuit at nearly the speed of light. This electric field exerts a force on all free electrons everywhere in the wire simultaneously. They all start drifting locally at the same time, causing an immediate flow of current throughout the circuit. It is the propagation of the electric field, not the physical travel of a single electron from the switch to the bulb, that turns the light on instantly.