For the circuit (Figure 1) , take E = 300V , R 1=4.0k? , R 2=6.0k? and assume th
ID: 1264679 • Letter: F
Question
For the circuit (Figure 1) , take E = 300V , R1=4.0k? , R2=6.0k? and assume the capacitor is initially uncharged. What are the currents in both resistors just after the switch is closed (in mA)? What is the voltage across the capacitor just after the switch is closed (in V)? What are the currents in both resistors a long time after the switch is closed (in mA)? What is the voltage across the capacitor a long time after the switch is closed (in V)? Long after the switch is closed it's again opened. What are I1 , I2 just after this switch opening (in mA)? What is Vc just after this switch opening (in V)? What are I1 , I2 a long time later (in mA)? What is Vc a long time later (in V)?
For the circuit (Figure 1) , take E = 300V , R1=4.0k? , R2=6.0k? and assume the capacitor is initially uncharged. What are the currents in both resistors just after the switch is closed (in mA)? What is the voltage across the capacitor just after the switch is closed (in V)? What are the currents in both resistors a long time after the switch is closed (in mA)? What is the voltage across the capacitor a long time after the switch is closed (in V)? Long after the switch is closed it's again opened. What are I1 , I2 just after this switch opening (in mA)? What is Vc just after this switch opening (in V)? What are I1 , I2 a long time later (in mA)? What is Vc a long time later (in V)?Explanation / Answer
a)
just after closing the switch , capacitor behaves as a short circuit ,
Hence , i(R1) = 300/4
i(R1) = 75 mA
i(R2) = 0 mA
b)
V(R1) = 300 V
as Capacitor is short
V(R2) = 0 V
c)
after a long time , capacitor is open circuited ,
I(R1) = I(R2)
I(R1) = 300/(4 + 6)
I(R1) = 30 mA
the current in both resistances is 30 mA
d)
V(R1) = 30 * 4
V(R1) = 120 V
and
V(R2) = 30 * 6
V(R2) = 180 V
e)
Here , Vc = V(R2)
Vc = 180 V
Now , I(R2) = 180/6
I(R2) = 30 mA
I(R1) = 0 mA
f)
later after a long time , the capacitor will discharge
Vc = 0 V