Question

Solve the following problem.

Three resistors 10 Ω, 20 Ω, and 30 Ω are connected in series combinations.

1. Find the equivalent resistance of series combination.
2. When this series combination is connected to 12V supply, by neglecting the value of internal resistance, obtain potential difference across each resistor.

Answer 1

Given: R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 30 Ω, V = 12 V

To find: i. Series equivalent resistance(R_s)
ii. Potential difference across each resistor (V₁, V₂, V₃)

Formula: i. R_s = R₁ + R₂ + R₃
ii. V = IR

Calculation: From formula (i),
R_s = 10 + 20 + 30 = 60 Ω
From formula (ii),
I = `"V"/"R"=12/60` = 0.2 A
∴ Potential difference across R₁,
V₁ = I × R₁ = 0.2 × 10 = 2 V
∴ Potential difference across R₂,
V₂ = 0.2 × 20 = 4 V
∴ Potential difference across R₃,
V₃ = 0.2 × 30 = 6 V

1. The equivalent resistance of the series combination is 60 Ω.
2. Potential difference across 10 Ω, 20 Ω, and 30 Ω resistors are 2 V, 4 V, and 6 V respectively.