Question

Two circuits, A and B, are connected in parallel to a 25 V battery, which has an internal resistance of 0.25 Ω. Circuit A consists of two resistors, 6 Ω and 4 Ω, connected in series. Circuit B consists of two resistors, 10 Ω and 5 Ω, connected in series. Determine the current flowing in, and the potential difference across, each of the four resistors. Also, find the power expended in the external circuit.

Answer 1

Explanation:

answer is as follows given in figure

Answer 2

The power expended in the external circuit is 100 J

The equivalent external resistance of the circuit will be

= (6+4) and (10+5)Ω in parallel

= 10Ω and 15 Ω in parallel

= $\frac{10 * 15}{10 +15}$ = 6Ω

Voltage of the battery = 25 V

So, current flowing in each circuit will be

I₄ = 25/10 = 2.5 A

I₆ = 25/10 = 2.5 A

I₁₀ = 25/15 = 1.67 A

I₅ = 25/15 = 1.67 A

Potential different across each resistors are

P₄ = 2.5 × 4 = 10 V

P₆ = 2.5 × 6 = 15 V

P₅ = 1.67 × 5 = 8.3 V

P₁₀ = 1.67 × 10 =16.7 V

Power expended in the external circuit is = $\frac{V^2}{R}$

                                                                     = $\frac{(25)^2}{(6+0.25)}$ J

                                                                     = 625/6.25 J

                                                                     = 100 J