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In the circuit shown in Fig. E26.41, both capacitors are initially charged to 45.0 V. (a) How long after closing the switch S will the potential across each capacitor be reduced to 10.0 V, and (b) what will be the current at that time?

Short Answer

Expert verified

Answer:

(a) After4.21×10-3s closing the switch the potential across each capacitor is reduced to .

(b) The at that time will be0.125A

Step by step solution

01

Concept

A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. The effect of a capacitor is known as capacitance.

02

Adding capacitance and resistance

From the above circuit we can see thatC1=15.0μF,C2=20.μF, and R1=30.0Ω,R2=50.0Ωthe switch is closed at the time .

Now, since both the capacitors are initially charged to we are required to find the equivalent capacitance and resistance to reduce the circuit to a simple circuit. Now we know that the capacitance adds in parallel and the resistance adds in series, hence we can write the thing in:

Req=R1+R2Ceq=C1+C2

Now let us substitute the values to get:

Req=(30.0+50.0)Ω=80.0ΩCeq=(15.0+20.0)μF=35.0μF

03

Calculating the time taken

While discharging the capacitor, the charge on the capacitor is given by:

Q=Q0e-tτ

Where τ=ReqCeqis the time constant, Q=CVCbut and Q=CVC0, we can write:

VC=VC0e-tτ

Now, we need to find how long after the closing of the switch S will the potential defense across each of the capacitors reduced to , therefore we need to solve for :

t=-τlnVCVC0

Now, putting the values we get:

t=-(2.80×10-3s)ln10.0V45.0V=4.21×10-3s

Therefore, the time taken is:

04

Calculating current

The current at a time is defined by

I=VC0Req=VC0Reqe-tτ

Plugging in the values we get:

I=VC0Reqe-tτ=45.080.0e-4.21×10-3s2.80×10-3s=0.125A

Therefore, the current is 0.125 A

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