Problem 12-42 Electric power - Part 3 - C

A \(1.10\;kW\) hairdryer and a \(60.0\;W\) light bulb are turned on in the same \(120\;V\; AC\) parallel circuit. Assuming three significant digits in the given voltage, determine the current in the (a) hairdryer (b) light bulb (c) fuse in the circuit. (d) If the fuse is rated at \(15 \;A,\) will it blow?

[Ans. (a) \(9.17\; A\)   (b) \(0.500\; A\)   (c) \(9.67 \;A\)  (d) \(\text{no}\)]

Accumulated Solution

Circuit diagram with a light and hairdryer.

\(W = \Delta VI \\ I_1 = W/ \Delta V = 1.1 \times 10^{-3}/120 = 9.17\; A\\ I_2 = W/ \Delta V = 60/120 = 0.50 \;A\)

Correct!
\(I_1 = W/ \Delta V = 1.1 \times 10^{-3}/120 = 9.17\; A\)
and for the light bulb
\(I_2 = W/ \Delta V = 60/120 = 0.50 \;A\)

 

The total current supplied by the power supply to the parallel circuit is:

(A)  \(1/I = 1/I_1 + 1/I_2\)

(B)   \(I = I_1 + I_2\)

(C)   \(I = I_1 I_2\)