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In Fig. C1=6.0μF,C2=3.0μF,C3=4.0μF, andC4=8.0μF.and C4=8.0μF.The capacitor network is connected to an applied potential difference Vab. After the charges on the capacitors have reached their final values, the voltage across C3 is 40.0 V. What are

(a) the voltages acrossC1andC2

(b) the voltage across C4.

(c) the voltage vabapplied to the network?

Short Answer

Expert verified

a) the voltages across C1andC2is 13.34V and 26.68V respectively.

b) the voltage across C4is 30.0 V.

c) the voltage Vab applied to the network is 70.0V.

Given:

C1=6.0μF,C2=3.0μF,C3=4.0μF,C4=8.0μF

Capacitors C1andC2are in series. The equivalent capacitance of 1 and 2C12is in parallel withC3.

C123andC4are in parallelV3=400V

Step by step solution

01

Calculate the voltage across C1 and C2.

We want to find V1andV2. In the parallel connection, the voltage does not change so the voltage across C1andC2is the same for C3so

V1+V2=40.0V (1)

In the series connection, the charge Q is the same for both capacitors C1andC2so we could obtain the next relation

Q1=Q2C1V1=C2V2V2=C1C2V1V2=6.0μF3.0μFV1V2=2V1

Let us plug this expression into equation (1) to get

V1+V2=40.0VV1+2V1=40.0VV1=13.34V

Now we can get the value ofV2by using the above equation

V2=2V1=2(13.34V)=26.68V

02

Calculate the voltage across c4.

Before finding v4we must calculate the equivalent capacitance for C1,C2andC3The equivalent capacitance for the three capacitors is C123and will be given by

C123=C12+C3C123=11C1+1C2+C3

Now let us put our values into this equation to get c123

C123=11C1+1C2+C3C123=116.0μF+13.0μF+4.0μFC123=6.0μF

Now we can conclude from the figure that C123andC4are in series, this means the charge is the same for C123andC4Hence, we could get the next relation to finding v4

Q123=Q4C123V3=C4V4V4=C123C4V3

Now we can put our values of C123,C4andV3into the above equation to get v4whereV3=V123=40.0V

V4=C123C4V3V4=6.0μF8.0μF40.0VV4=30V

03

Calculate the voltage Vab applied to the network.

We want to find the applied voltage across ab. As the capacitances C123andC4are in series, therefore, the applied voltage will be the summation of both voltage

Vab=V4+V123=30.0V+40.0V=70.0V

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