Question

calculate the energy transferred by a 5A current flowing through a resistor of 2 ohm for 30 minutes

Answer 1

Energy=power*time.
E=P*t.
Energy in joules.
Power in watts.
Time in seconds.
Power=(current)*(current)*resistor.
P=(I)^2*R.
P=(5*5)*2.
P=50 Watts.
Time=t=30mins=30*60seconds=1800secs.
Energy=E=P*t.
E=50*1800.
E=90000 joules.
E=90kilo joules.
E=90k joules.

Answer 2

Given :

Current (I) = 5 A

Resistance (R) = 2 ohm

Time (T) = 30 min

To Find :

The energy transferred by the current

Solution :

As time is given in minutes we need to first convert it into seconds.

T = 30 min = 30 × 60 = 1800 seconds

The amount of heat produced (H) = I²RT

                                                          = 5² × 2 × 1800

                                                          = 90000 J

Therefore, the energy transferred by a 5A current flowing through a resistor of 2 ohm for 30 minutes is 90000 J or 90 KJ