# Answer to Question #6445 in Electric Circuits for Rose

Question #6445

A coffee cup heater and a lamp are connected in parallel to the same 120 V outlet. Together, they use a total of 87 W of power. The resistance of the heater is 4.00 X 10^2Ω . Find the resistance of the lamp.

____________Ω

____________Ω

Expert's answer

Let R1=400Ω be a resistance of the heater, and R2 be the resistance of the

lamp,

I1, and I2 be respectively the currents through heater and lamp

respecively,

V=120V be the outlet, and P=87W be the total power which use the

heater and the lamp.

Since the heater and a lamp are connected in

parallel it follows that the total current is

I = I1+I2,

where

I1 =

U/R1

I2 = U/R2

On the other hand,

I = P/U

and thus

U/R1 + U/R2 = P/U

1/R1 + 1/R2 = P/U^2

1/R2 = P/U^2 -

1/R1 = (P R1-U^2)/(U^2 R1)

Hence

R2 = U^2 R1/(P R1-U^2)

=

= 120^2 * 400 / (87*400-120^2)

= 282.35 Ω

lamp,

I1, and I2 be respectively the currents through heater and lamp

respecively,

V=120V be the outlet, and P=87W be the total power which use the

heater and the lamp.

Since the heater and a lamp are connected in

parallel it follows that the total current is

I = I1+I2,

where

I1 =

U/R1

I2 = U/R2

On the other hand,

I = P/U

and thus

U/R1 + U/R2 = P/U

1/R1 + 1/R2 = P/U^2

1/R2 = P/U^2 -

1/R1 = (P R1-U^2)/(U^2 R1)

Hence

R2 = U^2 R1/(P R1-U^2)

=

= 120^2 * 400 / (87*400-120^2)

= 282.35 Ω

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