Question

1) 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 a 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, will the breaker interrupt his breakfast?

2) The student’s roommate wakes up and turns on the air conditioner. He realizes that the room is a mess, so starts to vacuum. Now does the circuit breaker interrupt breakfast?

Answer 1

1) The breaker will not interrupt the breakfast.

2) The breaker will interrupt the breakfast.

Step-by-step explanation:

1) We are given;

Average power of coffee maker; P_coff = 1200 W

Average power of microwave oven; P_mic = 1450 W

Average power of toaster; P_toast = 1146 W

Average power of frying pan; P_fryp = 1196 W

Effective voltage; v_eff = 120 V

current in circuit breaker; I_rms = 50A

Total average power;

ΣP1 = 1200 + 1450 + 1146 + 1196 = 4992 W

Let's find the effective current ;

I_eff = Total average power/Effective voltage

I_eff = ΣP1/V_eff

4992/120

I_eff = 41.6 A

This is less than the current of 50A in the circuit breaker. Thus, the breaker will not interrupt the breakfast.

2) we are told he turned on the air conditioner and also the vacuum cleaner.

From online sources;

Average power of air conditioner; P_aircon = 860 W

Average power of vacuum cleaner; P_vac = 630 W

Total sum of average power;

ΣP2 = ΣP1 + P_aircon + P_vac

ΣP2 = 4992 + 860 + 630

ΣP2 = 6482 W

Effective current is now;

I_eff = ΣP2/V_eff

I_eff = 6482/120

I_eff = 54.02 A

This is more than the current of 50A in the circuit breaker. Thus, the breaker will interrupt the breakfast.