Question

Study the following circuit and find: 

1. Effective resistance of the circuit
2. Current drawn from the battery
3. Potential difference across the 5 omega resistor

Answer 1

(i) Effective resistance of the circuit

R₃ and R₄ are in series and both are parallel to R₂

R₃ + R₄ = 10 Ohm

Effective Resistance across R₂ (R')

`1/("R'") = 1/"R"_2 + 1/("R"_3 + "R"_4)`

R' = 5 Ohm

Now, R₁', R' and R₅ are in series

Effective resistance of the circuit = R₁ + R' + R₅

= 5 + 5 + 10

= 20 Ohm

(ii) Current drawn from battery

V = IR

I = `"V"/"R"`

I = `20/20`

I = 1 A

(iii) Potential difference across 5-ohm resistor

V = IR

V = 1 × 5

V = 5V