Question

The arrangement of resistors shown in the above figure is connected to a battery.

The power dissipation in the 100 Ω resistor is 81 W. Calculate

1. the current in the circuit.
2. the reading in the voltmeter V₂.
3. the reading in the voltmeter V₁.

Answer 1

i. Power across the 100 Ω resistance = 81 W

P = I²R = 81 W

∴ I² = `81/100`

∴ I = `sqrt(81/100)`

= `9/10`

= 0.9 A

ii. Voltage across the 25 Ω resistors = V₂ = IR_eqv

For the 25 Ω resistors:

`1/R_"eqv" = 1/25 + 1/25`

= `2/25`

∴ R_eqv = `25/2`

= 12.5 Ω

∴ V₂ = 0.9 A × 12.5 Ω

= 11.25 V

iii. Voltage across 100 Ω = V₁₀₀ = IR

= 0.9 A × 100 Ω

= 90 V

∴ V₁ = 90 V + 11.25 V

= 101.25 V