An 1,820 W toaster, a 1,420 W electric frying pan, and a 55 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.) Required:
a. What current is drawn by each device?
b. Will this combination blow the 15-A fuse?

toaster- 15.1A

electric frying pan- 11.8 A

lamp- 0.5 A

b) The combination will blow the fuse.

Explanation:

When devices are connected in parallel, the potential difference across each of the devices is the same but the current through each is different. Hence;

V= 120 V

Power= IV

For the toaster;

I= 1820/120 = 15.1 A

For the electric frying pan;

I= 1420/120 = 11.8 A

For the lamp;

55/120 = 0.5 A

Total current = 15.1 +11.8 + 0.5 = 27.4 A

The combination will blow the fuse.

Explanation:    

step one:

Given data

power of toaster= 1,820 W  

power of electric frying pan= 1,420 W  

power of lamp= 55 W  

current of the outlet= 15 A

voltage of outlet = 120 V

step two

since all  three appliances are connected in parallel to the socket outlet, they will use the same voltage of 120 V and the currents will be different across each appliance,

Hence the current across the Toaster will be I₁

using P=I₁V we have

I₁= P/V

I₁= 1820/120 =  15.16 A

A. The current drawn by each device

the current across the  electric frying pan will be I₂

using P=I₂V we have

I₂= P/V

I₂= 1420/120 =  11.83 A

the current across the   lamp will be I₃

using P=I₃V we have

I₃= P/V

I₃= 55/120 =  0.45 A

therefore the total current drawn by all appliances will be

Total current = I₁+I₂+I₃= 15.16 +11.83+ 0.45= 27.44

B.  Will this combination blow the 15-A fuse?

27.44 A > 15 A by 45% ...and this will make fuse to blow

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i hope this you out, have a wonderful day

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.