An alternating current with a peak value of 5A passes through a resistor of resistance 10Ω. Calculate the rate at which energy is dissipated in the resistor

Gideonisco

5 years ago

The answer is B
R=10ohms
Peak value= 5/√2
Power =I²R

P =(5/√2)²×10

P=125watts

djsteech

11 years ago

I fink d ans is 125

P=Irms^2*R

where Irms=Io/root2

Irms=5/root2

(5/root2)^2 *10

=125w

Aderemi493

11 years ago

d ans is 125

Tim234D

1 year ago

Greetings, The Myschool Team.
The correct answer is B(125W) and not A(250W). In alternating current circuits, the current value used for determining energy dissipated is the root mean square (rms) value of current, not the peak current. This can be seen from the definition of rms current below:
"Root Mean Square (RMS) value of AC current is defined as the steady or DC current which when flowing through a circuit for a given time period produces the same heat as produced by the AC current flowing through the same circuit for the same time period. "
Source: https://electricalbaba.com/
RMS value= 5sin 45 = 5√2/2 = 3.54A
P = I²R = (5√2/2)² × 10
= 25/2 × 10
= 125W
Thus, the correct answer is B (125W) and not A(250W). Thank you.

tommy j

11 years ago

Pls hw did you get 20