A college student wakes up hungry He turns on the coffee mak

A college student wakes up hungry. He turns on the coffee maker (average wattage 1200 W), puts some oatmeal in the microwave oven (1450 W) to cook, puts couple of slices of bread in the toaster (1146 W), and starts making scrambled eggs in the electric frying pan (1196 W). All of these appliances in his dorm room are supplied by a 120 V (rms) branch circuit protected by a 50 A (rms) circuit breaker.

Solution

Total load(P) = 1200+1450+1146+1196 = 4992 watts

Voltage supply(V) = 120 V rms

All are heating loads. So the power factor approaches unity.

I*(pf)=P/V= 4992/120 = 41.6 amps

If pf is 1 the current will be 41.6 amps.

so the breaker would not trip.

If suppose power factor(pf) is 0.8 lagging

then I*0.8 =41.6 gives current I = 52 amps which would trip the circuit breaker