Question

. An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (The three devices are in parallel when plugged into the same socket.). (a) What current is drawn by each device? (b) Will this combination blow the 15-A fuse?

Answer 1

Toaster: 15A

Electric Frying Pan: 11.7A

Light bulb: 0.625A

(B) Yes the combination will blow the fuse.

Step-by-step explanation:

The foremula for calculating the power consumed by electrical appliances in a circuit is P = IV

The three appliances are connected in parallel which means that they will all have the same voltage but different current flowing through them.

The current I = P/V = power consumed by appliance / Voltage supplied to it.

The current drawn by:

the toaster = 1800/120 = 15A

the frying pan = 1400/120 = 11.7A

the light bulb = 75/120 = 0.625A

(B) To know if the combination of all three appliances will blow the fuse or not, we add all the current drawn by each of the electrical appliances. That ic the current drawn by the toaster, frying pan and the light bulb which is 15A + 11.7A + 0.625A = 27.3A. This current demand is greater than the fuse rating and as a result this combination of appliances will blow the fuse.

Answer 2

Step-by-step explanation:

The devices are connected in parallel, the same voltage will flow through them

1) I, current in the 1800 W toaster

P, power = I V

where p = 1800 W

1800 /120 = I

I drawn by the toaster = 15 A

2) I, current drawn by the 1400 W frying pan = 1400 / 120 = 11.667 A

3) I, current drawn by 75 W = 75 / 120 = 0.625 A

B) Total current drawn by the circuit = 15 A + 11.667 A + 0.625 A = 27.292 A which is greater than 15-A hence the combination will blow the fuse