Current Electricity Class 12 NCERT Solutions & PYQs
🔌 Chapter Overview
Welcome to Current Electricity. Is chapter mein hum charges ke flow, Ohm's law ke complex applications, aur Kirchhoff's circuit rules ko depth mein samjhenge. Ye chapter competitive exams (JEE/NEET) aur Boards ke numerical portion ke liye backbone hai.
Master Formulas:
Current $I = nAev_d$ | Resistance $R = \rho \frac{l}{A}$ | Power $P = \frac{V^2}{R}$
Current $I = nAev_d$ | Resistance $R = \rho \frac{l}{A}$ | Power $P = \frac{V^2}{R}$
📚 Part 1: Full NCERT Exercise Solutions
Q 3.1: The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is $0.4\ \Omega$, what is the maximum current that can be drawn from the battery?
Ans:
Given, Emf ($E$) = $12\ \text{V}$ and internal resistance ($r$) = $0.4\ \Omega$.
According to Ohm's law, $I = \frac{E}{R + r}$
For maximum current, external resistance $R$ must be $0$.
$I_{max} = \frac{E}{r} = \frac{12}{0.4} = \mathbf{30\ \text{A}}$.
Given, Emf ($E$) = $12\ \text{V}$ and internal resistance ($r$) = $0.4\ \Omega$.
According to Ohm's law, $I = \frac{E}{R + r}$
For maximum current, external resistance $R$ must be $0$.
$I_{max} = \frac{E}{r} = \frac{12}{0.4} = \mathbf{30\ \text{A}}$.
Q 3.2: A battery of emf 10 V and internal resistance $3\ \Omega$ is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Ans:
Given: $E = 10\ \text{V}$, $r = 3\ \Omega$, $I = 0.5\ \text{A}$.
1. **Finding External Resistance ($R$):**
$I = \frac{E}{R + r} \Rightarrow 0.5 = \frac{10}{R + 3}$
$R + 3 = \frac{10}{0.5} = 20 \Rightarrow R = 20 - 3 = \mathbf{17\ \Omega}$.
2. **Terminal Voltage ($V$):**
$V = E - Ir = 10 - (0.5 \times 3) = 10 - 1.5 = \mathbf{8.5\ \text{V}}$.
Given: $E = 10\ \text{V}$, $r = 3\ \Omega$, $I = 0.5\ \text{A}$.
1. **Finding External Resistance ($R$):**
$I = \frac{E}{R + r} \Rightarrow 0.5 = \frac{10}{R + 3}$
$R + 3 = \frac{10}{0.5} = 20 \Rightarrow R = 20 - 3 = \mathbf{17\ \Omega}$.
2. **Terminal Voltage ($V$):**
$V = E - Ir = 10 - (0.5 \times 3) = 10 - 1.5 = \mathbf{8.5\ \text{V}}$.
Q 3.3: (a) Three resistors $1\ \Omega, 2\ \Omega,$ and $3\ \Omega$ are combined in series. What is the total resistance? (b) If connected to 12 V battery, find potential drop across each.
Ans:
(a) Total resistance in series: $R_{eq} = R_1 + R_2 + R_3 = 1 + 2 + 3 = \mathbf{6\ \Omega}$.
(b) Current $I = \frac{V}{R_{eq}} = \frac{12}{6} = 2\ \text{A}$.
Potential drops: $V_1 = I R_1 = 2 \times 1 = \mathbf{2\ \text{V}}$; $V_2 = 2 \times 2 = \mathbf{4\ \text{V}}$; $V_3 = 2 \times 3 = \mathbf{6\ \text{V}}$.
(a) Total resistance in series: $R_{eq} = R_1 + R_2 + R_3 = 1 + 2 + 3 = \mathbf{6\ \Omega}$.
(b) Current $I = \frac{V}{R_{eq}} = \frac{12}{6} = 2\ \text{A}$.
Potential drops: $V_1 = I R_1 = 2 \times 1 = \mathbf{2\ \text{V}}$; $V_2 = 2 \times 2 = \mathbf{4\ \text{V}}$; $V_3 = 2 \times 3 = \mathbf{6\ \text{V}}$.
Q 3.4: Three resistors $2\ \Omega, 4\ \Omega,$ and $5\ \Omega$ are combined in parallel. What is the total resistance? If connected to 20 V, find current in each.
Ans:
Total resistance $\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} = \frac{10+5+4}{20} = \frac{19}{20}\ \Omega$.
$R_{eq} = \frac{20}{19} \approx \mathbf{1.05\ \Omega}$.
Current $I_1 = \frac{V}{R_1} = \frac{20}{2} = \mathbf{10\ \text{A}}$; $I_2 = \frac{20}{4} = \mathbf{5\ \text{A}}$; $I_3 = \frac{20}{5} = \mathbf{4\ \text{A}}$.
Total resistance $\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} = \frac{10+5+4}{20} = \frac{19}{20}\ \Omega$.
$R_{eq} = \frac{20}{19} \approx \mathbf{1.05\ \Omega}$.
Current $I_1 = \frac{V}{R_1} = \frac{20}{2} = \mathbf{10\ \text{A}}$; $I_2 = \frac{20}{4} = \mathbf{5\ \text{A}}$; $I_3 = \frac{20}{5} = \mathbf{4\ \text{A}}$.
Q 3.5: At room temperature ($27.0^\circ\text{C}$) the resistance of a heating element is $100\ \Omega$. What is the temperature if resistance is $117\ \Omega$? (Given $\alpha = 1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$)
Ans:
Using formula: $R_t = R_0[1 + \alpha(t - t_0)]$
$117 = 100[1 + 1.70 \times 10^{-4}(t - 27)]$
$1.17 - 1 = 1.70 \times 10^{-4}(t - 27)$
$0.17 = 1.70 \times 10^{-4}(t - 27) \Rightarrow t - 27 = \frac{0.17}{1.70 \times 10^{-4}} = 1000$
$t = 1000 + 27 = \mathbf{1027^\circ\text{C}}$.
Using formula: $R_t = R_0[1 + \alpha(t - t_0)]$
$117 = 100[1 + 1.70 \times 10^{-4}(t - 27)]$
$1.17 - 1 = 1.70 \times 10^{-4}(t - 27)$
$0.17 = 1.70 \times 10^{-4}(t - 27) \Rightarrow t - 27 = \frac{0.17}{1.70 \times 10^{-4}} = 1000$
$t = 1000 + 27 = \mathbf{1027^\circ\text{C}}$.
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🔥 Part 2: 5 Most Repeated PYQs (Board VIPs)
PYQ 2024, 2022 - 5 Marks
Q1: State Kirchhoff's Laws. Apply them to obtain the balance condition for a Wheatstone bridge.
Ans:
1. **Junction Rule:** Sum of currents entering a junction equals sum of currents leaving it ($\sum I = 0$).
2. **Loop Rule:** Algebraic sum of changes in potential around any closed loop is zero ($\sum \Delta V = 0$).
**Wheatstone Bridge:** For a balanced bridge, galvanometer current $I_g = 0$. Using Loop Rule in loops ABDA and BCDB, we get: $$\frac{P}{Q} = \frac{R}{S}$$
1. **Junction Rule:** Sum of currents entering a junction equals sum of currents leaving it ($\sum I = 0$).
2. **Loop Rule:** Algebraic sum of changes in potential around any closed loop is zero ($\sum \Delta V = 0$).
**Wheatstone Bridge:** For a balanced bridge, galvanometer current $I_g = 0$. Using Loop Rule in loops ABDA and BCDB, we get: $$\frac{P}{Q} = \frac{R}{S}$$
PYQ 2023 - 3 Marks
Q2: Define Drift Velocity and derive its relation with relaxation time.
Ans: Average velocity with which free electrons get drifted towards the positive terminal under an external electric field.
Acceleration $a = \frac{F}{m} = \frac{eE}{m}$.
Since $v = u + at$ and average initial velocity $u = 0$:
$$v_d = \frac{eE\tau}{m}$$ Where $\tau$ is the relaxation time.
Acceleration $a = \frac{F}{m} = \frac{eE}{m}$.
Since $v = u + at$ and average initial velocity $u = 0$:
$$v_d = \frac{eE\tau}{m}$$ Where $\tau$ is the relaxation time.
PYQ 2021 - 5 Marks
Q3: Explain the principle of a Potentiometer. Why is it superior to a voltmeter?
Ans: **Principle:** Potential drop across any portion of a uniform wire is directly proportional to its length ($V \propto l$), provided current is constant.
**Superiority:** A voltmeter draws some current from the source, so it measures terminal voltage. A potentiometer draws **no current** at the null point, thus measuring the exact Emf.
**Superiority:** A voltmeter draws some current from the source, so it measures terminal voltage. A potentiometer draws **no current** at the null point, thus measuring the exact Emf.
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⚡ Part 3: Extra CBT Concept Questions
- Define 1 Ampere in terms of charge flow.
- Why does resistance increase with temperature in metals?
- What is 'Specific Resistance' (Resistivity)? Write its SI unit.
- Can the terminal voltage of a cell be greater than its Emf? (Hint: Charging)
- Define 'Mobility' of electrons.
- What is the function of a 'Shunt' in a galvanometer?
- Why are standard resistors made of alloys like Manganin?
- What is the net charge in a current-carrying conductor?
- Define 'Ohmic' and 'Non-Ohmic' devices with examples.
- How does drift velocity change if potential difference is doubled?
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