Electricity - Resistance and Resistivity Solved Examples (Class 10 Physics)
Develop skills in series and parallel resistance calculations, equivalent resistance, and voltage-current distributions. These problems strengthen understa
TL;DR: Develop skills in series and parallel resistance calculations, equivalent resistance, and voltage-current distributions. These problems strengthen und…
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Develop skills in series and parallel resistance calculations, equivalent resistance, and voltage-current distributions. These problems strengthen understa
Electricity - Resistance and Resistivity — Solved Numerical Examples (Step by Step)
Example 1: Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in series. Find the total resistance.
Solution: In series, R_total = R1 + R2 + R3
R_total = 2 + 3 + 6 = 11 Ω
Example 2: Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel. Find the equivalent resistance.
Solution: In parallel, 1/R_total = 1/R1 + 1/R2 + 1/R3
1/R_total = 1/2 + 1/3 + 1/6
Finding common denominator (6):
1/R_total = 3/6 + 2/6 + 1/6 = 6/6 = 1
R_total = 1 Ω
Example 3: Two resistors of 4 Ω and 6 Ω are in parallel, and this combination is in series with a 2 Ω resistor. Find the total resistance.
Solution: First, find parallel resistance of 4 Ω and 6 Ω:
1/R_p = 1/4 + 1/6 = 3/12 + 2/12 = 5/12
R_p = 12/5 = 2.4 Ω
Total resistance = R_p + R_series = 2.4 + 2 = 4.4 Ω
Example 4: A 12 V battery is connected to a circuit with three resistors in series: 1 Ω, 2 Ω, and 3 Ω. Find the current through the circuit.
Solution: Total resistance R = 1 + 2 + 3 = 6 Ω
Using Ohm's law: V = IR
I = V/R = 12/6 = 2 A
Example 5: In a series circuit with a 24 V battery and total resistance of 6 Ω, find the current and power dissipated.
Solution: Current I = V/R = 24/6 = 4 A
Power P = VI = 24 × 4 = 96 W
Alternatively, P = I²R = 4² × 6 = 16 × 6 = 96 W
Example 6: A 10 Ω resistor and a 15 Ω resistor are in parallel. This combination is connected in series with a 5 Ω resistor and a 20 V battery. Find the current from the battery.
Solution: Parallel resistance: 1/R_p = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6
R_p = 6 Ω
Total resistance: R_total = 6 + 5 = 11 Ω
Current from battery: I = V/R = 20/11 = 1.82 A
Tips
- Series: R_total = R1 + R2 + R3 + ... (add resistances); same current through all.
- Parallel: 1/R_total = 1/R1 + 1/R2 + 1/R3 + ... (add reciprocals); same voltage across all.
- For two resistors in parallel: R_eq = (R1 × R2) / (R1 + R2) is a quick formula.
Frequently Asked Questions
Why is less current drawn when resistors are in series compared to parallel?
Series resistance is larger (resistances add), so total resistance increases, and by Ohm's law I = V/R, the current decreases.
What happens if one bulb burns out in a series vs. parallel circuit?
In series, the entire circuit breaks and all bulbs go off. In parallel, only that branch is broken; other branches remain lit.
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