Multiloop RC Circuit in the circuit shown below, the switch has been open for a
ID: 3164169 • Letter: M
Question
Multiloop RC Circuit in the circuit shown below, the switch has been open for a long time. At t = 0, the switch is closed. Just after the switch is closed, what is the current through each of the resistors? After the switch, has been closed for a very long time, what is the current through each of the resistors and the energy stored in each of the capacitors? At some time t, after the switch is closed, capacitor 4C is charged to half of its final charge. At that instant, what fraction of the power supplied by the battery is dissipated in resistor R? the switch has been closed a very long time and then is opened. What is the current through 3R just after the switch is opened?Explanation / Answer
a) Initially C is not charged, when switch is open
Right after the switch is closed, C acts as a wire
In that case, capacitor acts as a short circuit
For that equivalent resistance is
RT=6R
Current through each resistor is
I=Vs/6R
Since all the resistors are connected in series, current I will flow through all the resistors.
b) After long time switch is closed C is break in circuit
For that equivalent resistance is
RT=6R
Current through each resistor is
I=Vs/6R
Since all the resistors are connected in series, current I will flow through all the resistors.
To find energy:
Energy through the capacitor 4C is given as
E=(1/2)CV2
Here
E=(1/2)(4C)V12
Energy through the capacitor C is given as
E=(1/2)(C)V22
To find V1 and V2 , apply voltage divider rule
V2= V1 x(2R/(3R+2R))
= V1 (2R/5R)
V2 = 2V1/5
V1= Vsx(5R/(R+5R))
= V1 (5R/6R)
V1 = 5Vs/6
=> V2 = ( 2(5Vs/6))/5
V2 = Vs/3
Now energy through the capacitor 4C is given as
E =(1/2)(4C) (5Vs/6)2
=25CVs2/18
Energy through the capacitor C is given as
E=(1/2)(C) (Vs/3)2
= CVs2/18
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