Current Electricity Solved Examples (Class 12 Physics)
Current electricity deals with the flow of charge and the forces that create and oppose this flow. These examples cover Ohm's law, resistivity, power dissi
TL;DR: Current electricity deals with the flow of charge and the forces that create and oppose this flow. These examples cover Ohm's law, resistivity, power…
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Current electricity deals with the flow of charge and the forces that create and oppose this flow. These examples cover Ohm's law, resistivity, power dissi
Current Electricity — Solved Numerical Examples (Step by Step)
Example 1: A resistor of 10 ohms carries a current of 2 A. Calculate the voltage across it and power dissipated.
Solution: Given: Resistance R = 10 Ω, current I = 2 A
Using Ohm's law: V = IR
V = 2 × 10 = 20 V
Power dissipated: P = I²R = 2² × 10 = 4 × 10 = 40 W
Alternatively: P = VI = 20 × 2 = 40 W
Or: P = V²/R = 20²/10 = 400/10 = 40 W
Example 2: A copper wire of length 2 m and cross-sectional area 1 mm² has resistivity 1.7 × 10⁻⁸ Ω m. Calculate its resistance.
Solution: Given: Length L = 2 m, area A = 1 mm² = 1 × 10⁻⁶ m², resistivity ρ = 1.7 × 10⁻⁸ Ω m
Using R = ρL/A
R = (1.7 × 10⁻⁸ × 2) / (1 × 10⁻⁶)
R = (3.4 × 10⁻⁸) / (1 × 10⁻⁶)
R = 3.4 × 10⁻⁸⁺⁶ = 3.4 × 10⁻² = 0.034 Ω
Example 3: Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel. Find the equivalent resistance.
Solution: Given: R1 = 2 Ω, R2 = 3 Ω, R3 = 6 Ω
For parallel combination: 1/R_eq = 1/R1 + 1/R2 + 1/R3
1/R_eq = 1/2 + 1/3 + 1/6
1/R_eq = 3/6 + 2/6 + 1/6 = 6/6 = 1
R_eq = 1 Ω
Example 4: Two cells of EMF 1.5 V and internal resistance 0.5 Ω each are connected in series and attached to an external resistance of 10 Ω. Calculate the current and terminal voltage of each cell.
Solution: Given: EMF of each cell E = 1.5 V, internal resistance r = 0.5 Ω each, external resistance R = 10 Ω
Total EMF = 1.5 + 1.5 = 3 V
Total internal resistance = 0.5 + 0.5 = 1 Ω
Current: I = Total EMF / (Total R + Total r)
I = 3 / (10 + 1) = 3/11 ≈ 0.27 A
Terminal voltage of one cell: V = E - Ir
V = 1.5 - (3/11) × 0.5 = 1.5 - 1.5/11 = 1.5(1 - 1/11) = 1.5 × 10/11 ≈ 1.36 V
Example 5: A heating element rated 1000 W, 220 V is used for 5 hours. Calculate the cost of electricity if the rate is Rs 5 per unit (kilowatt-hour).
Solution: Given: Power P = 1000 W = 1 kW, voltage V = 220 V, time t = 5 hours, cost rate = Rs 5/kWh
Energy consumed: E = P × t = 1 kW × 5 h = 5 kWh (5 units)
Cost = Energy × Rate = 5 × 5 = Rs 25
Example 6: A potential difference of 100 V is applied across a resistor of 50 Ω. Calculate the current, power dissipated, and energy dissipated in 30 minutes.
Solution: Given: Voltage V = 100 V, resistance R = 50 Ω, time t = 30 min = 1800 s
Current: I = V/R = 100/50 = 2 A
Power: P = V²/R = 100²/50 = 10000/50 = 200 W
Alternatively: P = VI = 100 × 2 = 200 W
Energy: E = Pt = 200 × 1800 = 360000 J = 360 kJ
Or in kWh: E = (200 W / 1000) × (30/60) h = 0.2 × 0.5 = 0.1 kWh
Tips
- Always convert power ratings to watts and time to appropriate units (seconds or hours) before calculating energy.
- For series circuits, voltage adds but current remains same. For parallel circuits, current adds but voltage remains same.
- Use Ohm's law (V = IR) to find missing quantities in any circuit problem.
- Remember that power dissipation in a resistor increases with the square of current (P = I²R) or voltage (P = V²/R).
Frequently Asked Questions
What is the difference between EMF and terminal voltage?
EMF (electromotive force) is the total energy provided by a cell per unit charge and is constant for a cell. Terminal voltage is the voltage available across the external circuit and is less than EMF because of the voltage drop across the internal resistance (V_terminal = EMF - Ir).
Why is a unit called kilowatt-hour?
A kilowatt-hour (kWh) is the energy consumed when a device of power 1000 watts operates for 1 hour. It is used for commercial electricity billing because it represents practical quantities. 1 kWh = 3.6 million joules (3.6 × 10⁶ J).
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