How is the total current found from the 90 V across C = 5 ohm?

I = 90/5 = 18 A, since C carries the total current

12.5 ohm, since I_B = 18 - 6 = 12 A and V = 150 V

2073 Chaitra · 6 marks

Resistors A, B in parallel with C in series

Short answer

Total current I = 90V/5Ω = 18A. Applied voltage V = 4320W/18A = 240V. Voltage across parallel pair = 240V − 90V = 150V. I_A = 150V/25Ω = 6A, I_B = 18A − 6A = 12A. R_B = 150V/12A = 12.5Ω. (i) 12.5Ω, (ii) 240V, (iii) I_A=6A, I_B=12A, I_C=18A.

A direct current circuit comprises two resistors A of value 25Ω and B of unknown value, connected in parallel together with a third resistor C of value 5Ω connected in series with the parallel group. The potential difference across C is found to be 90V. If the total power in the circuit is 4320 watt. Calculate (i) the value of resistor B, (ii) the voltage applied to the ends of the whole circuit, (iii) the current in each resistor.

A (25 Ω) and B are in parallel; this group is in series with C (5 Ω).

 o--+--[A 25]--+--[C 5]--o
    +--[B ?]---+

Step 1: Total current

I = V_C/R_C = 90/5 = 18 A

Step 2: Applied voltage

P = V I, so V = 4320/18 = 240 V

Step 3: Voltage across the parallel group

V_p = 240 − 90 = 150 V

Step 4: Currents

I_A = 150/25 = 6 A
I_B = 18 − 6 = 12 A
I_C = 18 A

Step 5: Resistor B

R_B = 150/12 = 12.5 Ω

(Check: 18² × 5 + 6² × 25 + 12² × 12.5 = 1620 + 900 + 1800 = 4320 W.)

Answer: (i) R_B = 12.5 Ω; (ii) V = 240 V; (iii) I_A = 6 A, I_B = 12 A, I_C = 18 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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