Circuit A in a house has a voltage of 208 V and is limited by a 40.0-A circuit breaker. Circuit B is at 120.0 V and has a 20.0-A circuit breaker. What is the ratio of the maximum power delivered by circuit A to that delivered by circuit B

Question

Physics

Keaton Owen

23 October, 12:16

Circuit A in a house has a voltage of 208 V and is limited by a 40.0-A circuit breaker. Circuit B is at 120.0 V and has a 20.0-A circuit breaker. What is the ratio of the maximum power delivered by circuit A to that delivered by circuit B

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Answers (1)

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Carina Cross · Answer 1

Given Information:

Voltage of circuit A = Va = 208 Volts

Current of circuit A = Ia = 40 Amps

Voltage of circuit B = Vb = 120 Volts

Current of circuit B = Ib = 20 Amps

Required Information:

Ratio of power = Pa/Pb = ?

Answer:

Ratio of power = Pa/Pb = 52/15

Explanation:

Power can be calculated using Ohm's law

P = VI

Where V is the voltage and I is the current flowing in the circuit.

The power delivered by circuit A is

Pa = Va*Ia

Pa = 208*40

Pa = 8320 Watts

The power delivered by circuit B is

Pb = Vb*Ib

Pb = 120*20

Pb = 2400 Watts

Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is

Pa/Pb = 8320/2400

Pa/Pb = 52/15