A 50 volt battery is connected in series with a 20 omega res

A 5.0 volt battery is connected in series with a 20 omega resister in a closed-loop circuit. (a) What is the current in the loop in steady state? (b) What is the voltage drop arrow the resistor? (c) At what rate is heat dissipated by the resistor?

Solution

V = 5.0 V

R = 20 ohm

ohm\'s law

V = iR

A. i = V/R

current i = 5/20 = 1/4

current = 0.25 Amp.

B. Voltage drop across Resistor will be

V = i*R = 0.25*20

V = 5 V

C. Heat dissipated

Power = V*i = 5*0.25 = 1.25 W