Answer to Question #197701 in Electric Circuits for Plea

Question

A 240-V clothes dryer draws a current of 15 A. How much energy, in kilowatt-hours and in joules, does it use in 45 minutes of operation?

Answer 1

Gives

Voltage (v)=240V

Current (i)=15A

Time=45min=0.75h

We know that

$P=\frac{E}{t}$

$E={P}\times{t}$

Where P=power

E=energy

t=time

When

P$=IV$

Put value

$P=15\times240=3600W$

$E=3600\times0.75Wh$

$E=2700Wh$

Energy found kWh

$E=\frac{2700}{1000}=2.7KWh$

1KWh=$3.6\times10^6J$

E=$2.7\times3.6\times10^{6}J$

$E=9.72\times20^6J$