# 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 10^2&Omega; . Find the resistance of the lamp. ____________&Omega;
Let R1=400&Omega; 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 &Omega;

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