Problem: Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 × 10⁻⁷ m² carrying a current of 2.0 A. Assume that each copper atom contributes one conduction electron. Density of copper = 8.96 × 10³ kg/m³, atomic mass = 63.5 u, Avogadro’s number = 6.022 × 10²³ mol⁻¹.

Solution 3 Marks

Given:
$A = 2.5 \times 10^{-7}\, \text{m}^2,\; I = 2.0\, \text{A},\; \rho_m = 8.96 \times 10^3\, \text{kg/m}^3$
Atomic mass $M = 63.5\, \text{g/mol} = 63.5 \times 10^{-3}\, \text{kg/mol}$
$N_A = 6.022 \times 10^{23}\, \text{mol}^{-1},\; e = 1.6 \times 10^{-19}\, \text{C}$

Step 1: Calculate electron density (n)
Number of atoms per m³ = $\frac{\rho_m N_A}{M} = \frac{(8.96 \times 10^3) \times (6.022 \times 10^{23})}{63.5 \times 10^{-3}}$
$n = 8.5 \times 10^{28}\, \text{electrons/m}^3$ (assuming 1 conduction electron per atom)

Step 2: Apply drift velocity formula
$v_d = \frac{I}{n e A} = \frac{2.0}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2.5 \times 10^{-7})}$
$v_d = \frac{2.0}{3.4 \times 10^3} = 5.88 \times 10^{-4}\, \text{m/s} = 0.588\, \text{mm/s}$

Comparison:
• Thermal speed at room temperature ≈ 10⁵ m/s
• Drift speed is ~10⁻⁹ times smaller than thermal speed
• Electric field propagation speed = 3 × 10⁸ m/s (speed of light)

$\boxed{v_d = 0.588\,\text{mm/s}}$

2. Resistance and Resistivity

Problem: A wire of length 2.0 m and diameter 1.0 mm has a resistance of 50 Ω at 20°C. Calculate (a) the resistivity of the material, and (b) the resistance of a wire of the same material with length 3.0 m and diameter 0.5 mm at the same temperature.

Solution 2 Marks

Given (Part a):
$l_1 = 2.0\, \text{m},\; d_1 = 1.0\, \text{mm} = 10^{-3}\, \text{m},\; R_1 = 50\, \Omega$
Area $A_1 = \pi r_1^2 = \pi (0.5 \times 10^{-3})^2 = 7.854 \times 10^{-7}\, \text{m}^2$

Part (a): Resistivity
$R = \rho \frac{l}{A} \Rightarrow \rho = \frac{R A}{l} = \frac{50 \times 7.854 \times 10^{-7}}{2.0} = 1.96 \times 10^{-5}\, \Omega\cdot\text{m}$

Part (b): New resistance
$l_2 = 3.0\, \text{m},\; d_2 = 0.5\, \text{mm} = 0.5 \times 10^{-3}\, \text{m}$
$A_2 = \pi (0.25 \times 10^{-3})^2 = 1.9635 \times 10^{-7}\, \text{m}^2$
$R_2 = \rho \frac{l_2}{A_2} = (1.96 \times 10^{-5}) \times \frac{3.0}{1.9635 \times 10^{-7}} = 300\, \Omega$

$\boxed{\text{(a) } \rho = 1.96 \times 10^{-5}\,\Omega\cdot\text{m} \quad \text{(b) } R_2 = 300\,\Omega}$

3. Temperature Dependence of Resistance

Problem: The resistance of a nichrome wire at 0°C is 20 Ω. What is its resistance at 200°C? Temperature coefficient of resistance of nichrome = 1.7 × 10⁻⁴ °C⁻¹.

Solution 2 Marks

Given:
$R_0 = 20\, \Omega,\; T_0 = 0^\circ\text{C},\; T = 200^\circ\text{C},\; \alpha = 1.7 \times 10^{-4}\, ^\circ\text{C}^{-1}$

Apply temperature-resistance relation:
$R_T = R_0 [1 + \alpha (T - T_0)]$
$R_{200} = 20 [1 + (1.7 \times 10^{-4}) \times 200]$
$R_{200} = 20 [1 + 0.034] = 20 \times 1.034 = 20.68\, \Omega$

$\boxed{R_{200^\circ\text{C}} = 20.68\,\Omega}$

4. Electrical Power and Energy Loss

Problem: An electric heater rated 1000 W, 220 V is connected to a 220 V supply. Calculate (a) the resistance of the heater coil, (b) the current drawn, and (c) the energy consumed in 2 hours. If the same heater is connected to a 110 V supply, what will be the power consumed?

Solution 3 Marks

Given:
Rated power $P = 1000\, \text{W}$ , rated voltage $V = 220\, \text{V}$

Part (a): Resistance
$P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P} = \frac{(220)^2}{1000} = \frac{48400}{1000} = 48.4\, \Omega$

Part (b): Current
$I = \frac{P}{V} = \frac{1000}{220} = 4.545\, \text{A}$
(Alternatively: $I = \frac{V}{R} = \frac{220}{48.4} = 4.545\, \text{A}$ )

Part (c): Energy consumed
Time $t = 2\, \text{hours} = 7200\, \text{s}$
Energy $E = P \times t = 1000 \times 7200 = 7.2 \times 10^6\, \text{J} = 2\, \text{kWh}$

Part (d): Power at 110 V
Resistance remains constant (48.4 Ω)
$P' = \frac{(V')^2}{R} = \frac{(110)^2}{48.4} = \frac{12100}{48.4} = 250\, \text{W}$
Note: Power reduces to 1/4th when voltage is halved (since P ∝ V²)

$\boxed{\text{(a) } 48.4\,\Omega \quad \text{(b) } 4.545\,\text{A} \quad \text{(c) } 2\,\text{kWh} \quad \text{(d) } 250\,\text{W}}$

5. Cell with Internal Resistance

Problem: A cell of emf 2.0 V and internal resistance 0.5 Ω is connected to a resistor of resistance 4.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the cell, and (c) the power dissipated in the external resistor.

Circuit diagram showing cell with internal resistance connected to external resistor — Simple circuit with cell (emf ε, internal resistance r) connected to external resistor R

Solution 2 Marks

Given:
$\varepsilon = 2.0\, \text{V},\; r = 0.5\, \Omega,\; R = 4.5\, \Omega$

Part (a): Current
Total resistance = $R + r = 4.5 + 0.5 = 5.0\, \Omega$
$I = \frac{\varepsilon}{R + r} = \frac{2.0}{5.0} = 0.4\, \text{A}$

Part (b): Terminal voltage
$V = \varepsilon - I r = 2.0 - (0.4 \times 0.5) = 2.0 - 0.2 = 1.8\, \text{V}$
(Alternatively: $V = I R = 0.4 \times 4.5 = 1.8\, \text{V}$ )

Part (c): Power dissipated
$P = I^2 R = (0.4)^2 \times 4.5 = 0.16 \times 4.5 = 0.72\, \text{W}$

$\boxed{\text{(a) } 0.4\,\text{A} \quad \text{(b) } 1.8\,\text{V} \quad \text{(c) } 0.72\,\text{W}}$

6. Cells in Series Combination

Problem: Three identical cells, each of emf 1.5 V and internal resistance 0.2 Ω, are connected in series to an external resistor of 5.4 Ω. Calculate (a) the equivalent emf and internal resistance of the combination, and (b) the current flowing in the circuit.

Solution 2 Marks

Given:
$\varepsilon_1 = \varepsilon_2 = \varepsilon_3 = 1.5\, \text{V},\; r_1 = r_2 = r_3 = 0.2\, \Omega,\; R = 5.4\, \Omega$

Part (a): Equivalent parameters
For series combination:
$\varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2 + \varepsilon_3 = 1.5 + 1.5 + 1.5 = 4.5\, \text{V}$
$r_{\text{eq}} = r_1 + r_2 + r_3 = 0.2 + 0.2 + 0.2 = 0.6\, \Omega$

Part (b): Current
$I = \frac{\varepsilon_{\text{eq}}}{R + r_{\text{eq}}} = \frac{4.5}{5.4 + 0.6} = \frac{4.5}{6.0} = 0.75\, \text{A}$

$\boxed{\text{(a) } \varepsilon_{\text{eq}} = 4.5\,\text{V},\; r_{\text{eq}} = 0.6\,\Omega \quad \text{(b) } 0.75\,\text{A}}$

7. Cells in Parallel Combination

Problem: Two identical cells, each of emf 2.0 V and internal resistance 0.4 Ω, are connected in parallel to an external resistor of 3.8 Ω. Calculate the current flowing through the external resistor.

Solution 2 Marks

Given:
$\varepsilon_1 = \varepsilon_2 = 2.0\, \text{V},\; r_1 = r_2 = 0.4\, \Omega,\; R = 3.8\, \Omega$

Step 1: Equivalent parameters for parallel cells
For n identical cells in parallel:
$\varepsilon_{\text{eq}} = \varepsilon = 2.0\, \text{V}$ (same emf)
$r_{\text{eq}} = \frac{r}{n} = \frac{0.4}{2} = 0.2\, \Omega$

Step 2: Total current
$I = \frac{\varepsilon_{\text{eq}}}{R + r_{\text{eq}}} = \frac{2.0}{3.8 + 0.2} = \frac{2.0}{4.0} = 0.5\, \text{A}$

Verification:
Current through each cell = $I/2 = 0.25\, \text{A}$
Terminal voltage of each cell = $\varepsilon - (I/2)r = 2.0 - (0.25 \times 0.4) = 1.9\, \text{V}$
Voltage across R = $I R = 0.5 \times 3.8 = 1.9\, \text{V}$ ✓

$\boxed{I = 0.5\,\text{A}}$

8. Kirchhoff’s Rules Application

Problem: Determine the current in each branch of the network shown below. The circuit has a 10 V battery and resistors of 1 Ω, 2 Ω, and 3 Ω arranged as shown.

Circuit diagram with three resistors and a battery requiring Kirchhoff's rules application — Network requiring application of Kirchhoff’s junction and loop rules

Solution 3 Marks

Step 1: Assign currents and apply junction rule
Let currents be $I_1$ (through 1 Ω), $I_2$ (through 2 Ω), and $I_3$ (through 3 Ω)
At junction: $I_1 = I_2 + I_3$ …(1)

Step 2: Apply loop rule to left loop (clockwise)
$10 - I_1(1) - I_2(2) = 0$
$I_1 + 2I_2 = 10$ …(2)

Step 3: Apply loop rule to right loop (clockwise)
$-I_3(3) + I_2(2) = 0$
$2I_2 = 3I_3 \Rightarrow I_3 = \frac{2}{3}I_2$ …(3)

Step 4: Solve equations
Substitute (3) in (1): $I_1 = I_2 + \frac{2}{3}I_2 = \frac{5}{3}I_2$
Substitute in (2): $\frac{5}{3}I_2 + 2I_2 = 10 \Rightarrow \frac{11}{3}I_2 = 10$
$I_2 = \frac{30}{11} = 2.727\, \text{A}$
$I_3 = \frac{2}{3} \times \frac{30}{11} = \frac{20}{11} = 1.818\, \text{A}$
$I_1 = \frac{5}{3} \times \frac{30}{11} = \frac{50}{11} = 4.545\, \text{A}$

$\boxed{I_1 = 4.545\,\text{A},\; I_2 = 2.727\,\text{A},\; I_3 = 1.818\,\text{A}}$

9. Wheatstone Bridge Balance Condition

Problem: In a Wheatstone bridge experiment, four resistances are arranged as P = 10 Ω, Q = 20 Ω, R = 15 Ω, and S (unknown). The bridge is balanced when S is adjusted to a certain value. Calculate the value of S. If S is replaced by two resistors 30 Ω and X in parallel, and the bridge remains balanced, find X.

Solution 2 Marks

Given:
$P = 10\, \Omega,\; Q = 20\, \Omega,\; R = 15\, \Omega$

Part (a): Balance condition
For balanced Wheatstone bridge:
$\frac{P}{Q} = \frac{R}{S} \Rightarrow S = R \times \frac{Q}{P} = 15 \times \frac{20}{10} = 30\, \Omega$

Part (b): Parallel combination
Equivalent resistance of 30 Ω and X in parallel:
$S_{\text{eq}} = \frac{30 \times X}{30 + X}$
For balance: $S_{\text{eq}} = 30\, \Omega$
$\frac{30X}{30 + X} = 30 \Rightarrow 30X = 900 + 30X$
This gives no finite solution → Bridge cannot remain balanced with this arrangement.
Correction: If bridge remains balanced with new arrangement, then:
$\frac{30X}{30+X} = 30 \Rightarrow$ impossible unless X → ∞ (open circuit)
Practical interpretation: For balance to be maintained, the parallel combination must still equal 30 Ω, which requires X → ∞ (i.e., effectively removing the second resistor).

$\boxed{\text{(a) } S = 30\,\Omega \quad \text{(b) Not possible for finite } X}$

10. Meter Bridge Experiment

Problem: In a meter bridge experiment, the null point is obtained at 40 cm from one end when an unknown resistance X is connected in the left gap and a standard resistance of 15 Ω is connected in the right gap. Calculate the value of X. If the resistances are interchanged, find the new balance point.

Meter bridge setup showing wire, gaps, and galvanometer for resistance measurement — Meter bridge arrangement for determining unknown resistance

Solution 2 Marks

Given:
Balance length from left end $l_1 = 40\, \text{cm}$
Standard resistance $R = 15\, \Omega$ (in right gap)
Unknown resistance $X$ (in left gap)

Part (a): Calculate X
For meter bridge:
$\frac{X}{R} = \frac{l_1}{100 - l_1} = \frac{40}{60} = \frac{2}{3}$
$X = R \times \frac{2}{3} = 15 \times \frac{2}{3} = 10\, \Omega$

Part (b): After interchanging resistances
Now X = 10 Ω in right gap, R = 15 Ω in left gap
Let new balance point be at $l_2$ cm from left end:
$\frac{15}{10} = \frac{l_2}{100 - l_2} \Rightarrow \frac{3}{2} = \frac{l_2}{100 - l_2}$
$3(100 - l_2) = 2l_2 \Rightarrow 300 = 5l_2 \Rightarrow l_2 = 60\, \text{cm}$
Note: Balance points are complementary (40 cm and 60 cm add to 100 cm)

$\boxed{\text{(a) } X = 10\,\Omega \quad \text{(b) } 60\,\text{cm from left end}}$

Exam Tip: For numerical problems on Current Electricity:

• Always distinguish between emf (ε) and terminal voltage (V = ε – Ir)
• For series resistors: R_eq = R₁ + R₂ + … ; same current flows through all
• For parallel resistors: 1/R_eq = 1/R₁ + 1/R₂ + … ; same voltage across all
• When applying Kirchhoff’s rules: (i) Junction rule for currents, (ii) Loop rule for voltages
• For temperature dependence: R_T = R₀[1 + α(T – T₀)] — valid for limited temperature ranges
• Power calculations: P = VI = I²R = V²/R — choose formula based on known quantities

← Back to Theory Next: Moving Charges & Magnetism →

10 Solved Numericals on Current Electricity

10 Solved Numericals on Current Electricity

1. Drift Velocity of Electrons

2. Resistance and Resistivity

3. Temperature Dependence of Resistance

4. Electrical Power and Energy Loss

5. Cell with Internal Resistance

6. Cells in Series Combination

7. Cells in Parallel Combination

8. Kirchhoff’s Rules Application

9. Wheatstone Bridge Balance Condition

10. Meter Bridge Experiment

Editorial Team

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1. Drift Velocity of Electrons

2. Resistance and Resistivity

3. Temperature Dependence of Resistance

4. Electrical Power and Energy Loss

5. Cell with Internal Resistance

6. Cells in Series Combination

7. Cells in Parallel Combination

8. Kirchhoff’s Rules Application

9. Wheatstone Bridge Balance Condition

10. Meter Bridge Experiment

Editorial Team

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