Please help me to solve this problem... In the circuit below R = 115 omega and t
ID: 1616096 • Letter: P
Question
Please help me to solve this problem...
In the circuit below R = 115 omega and the capacitor is uncharged when the switch s is open. The switch is then closed at time t = 0 s. At t = 0 s (a) what is the voltage drop across the capacitor, (b) what is the charge on the capacitor, and (c) what is the current through R1?After a VERY long time after the switch has been closed (d) what is the charge on the capacitor (e) what is the current through R_1 (f) what is the current through R_2 nd (g) what is the current through R_3?Explanation / Answer
This is very-much similar to a Charging R-C Circuit.
At t = 0 s, after the Switch S is closed, the Capacitor, C acts as a Short Circuit instantaneously.
Max. Charging Current flows across the Capacitor via R1
a) Voltage Drop across the Capacitor at t = 0s,Vc(0) = 0V ( Capacitor acts as Short Circuit)
b) As the charge on the Capacitor, C is given by q = CVc , where Vc is the voltage across the Capacitor,
we have q (t=0) = C * Vc(0) = 0 C.
c) Current through R1 = 22v/ R = 22/115 = 0.191A
d) After a long time, The circuit must have attained steady state.
The Capacitor must have been charged and will act as an Open Circuit.
Vc ( at t = infinity) = 22V * ( 2R II 3R) / ( ( 2R II 3R) + R)
= 22V * ( 1.2R)/ ((1.2R) + R)
= 22V * (1.2R)/(2.2R)
= 12V
e) Current Through R1= (22-12)/R
= 10/115 = 0.0869 A
f) Current through R2= 12V/ 2R
= 12/(2*115) = 12/230 = 0.0521A
g) Current through R3= 12V/ 3R
= 12/(3*115) = 12/ 345 = 0.0347A