A capacitor is connected in the circuit shown with two resistors, A and B. Resis
ID: 1594338 • Letter: A
Question
A capacitor is connected in the circuit shown with two resistors, A and B. Resistor A has a resistance of 2R, while resistor B has a resistance of R. The emf of the battery, which has negligible internal resistance, is 12.0 V. Immediately after the switch is closed, the capacitor has a potential difference of 6.00 V, with the top plate having a higher potential, and the current through resistor A is 2.60 A. At that instant in time, calculate the following. the resistance of resistor A the current through resistor B the current through the capacitor A long time after the switch is closed, the current through resistor A has reached a stable value. What is that value?Explanation / Answer
Ra = 2*R
Rb = R
E = 12.0 V
When switch is closed -
E - I*Ra = 6.0
12.0 - 2.6* Ra = 6.0
Ra = 2.31 ohm
Rb = 2.31/2
Rb = 1.15 ohm
Current through Capacitor, I = 6.0/1.15 - 2.6
Current through Capacitor, I = 2.6 A
a long time after the switch is closed,
I = V/(Ra+Rb)
I = 12.0/(2.31+1.15)
I = 3.46 A
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.