Problem 12-42 Electric power - Part 4 - B

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!

\(I = I_1 + I_2 = 9.17 + 0.50 = 9.67 \;A\)

This is the current out of the power supply and through the fuse. If it is rated at \(15\; A\) it will not blow.

 

You have completed this problem.