Four capacitors are arranged in the circuit shown in the figure. The capacitors

Question

Four capacitors are arranged in the circuit shown in the figure. The capacitors have the values C1 = 31.5 ?F, C2 = 13.5 ?F, C3 = 45.5 ?F, C4 = 85.5 ?F, and the power supply is set at voltage V = 21.5 V. What is the equivalent capacitance of the circuit? What is the charge on capacitor C3? What is the potential difference across capacitor C4? What is the potential energy stored in capacitor C2?

The first answer of 40.785 IS CORRECT. I just need the other answers.

sapling learning Four capacitors are arranged in the circuit shown in the figure. The capacitors have the values C,-31.5 lf, C2 = 13.5 C3 = 45.5 F, C4 85.5 LF, and the power supply is set at voltage V 21.5 V C. What is the equivalent capacitance of the circuit? Number 40.785 What is the charge on capacitor C3? Number What is the potential difference across capacitor C4? Number What is the potential energy stored in capacitor C2? Number

Explanation / Answer

Given that

The capacitors have the values C1 = 31.5 F, C2 = 13.5 F, C3 = 45.5 F, C4 = 85.5 F

The power supply is set at voltage V = 21.5 V.

From the given diagram

C2,C3 & C4 are in series than equivalent capacitance is given by

1/C234 =1/C2+1/C3+1/C4

           =1/13.5+1/45.5+1/85.5

           =0.07407+0.0219+0.01169=0.10766

C234 =9.2885uF

Now C1 and C234 are in parallel than the equivalent capacitance is given by

Ceq =C1+C234 =31.5uF+9.2885uF =40.7885uF

Now the total charge in the circuit is given by

Q =Ceq*V =(40.7885uF)(21.5V) =876.952*10-6C

Now in C234 and C1 are in parallel than voltage remians the same but charge will be different

Therefore

   Then charge on capacitor C1 is Q1 =C1*V =(31.5uF)(21.5) =677.25uC

Now the charge on equivalent capacitance C234 is Q234 =C234*V =(9.2885uF)(21.5V )=199.702uC

Now C2 ,C3 and C4 are in series than the charge remains the same on all the three capacitors but the voltage will be varying

So the the charge on C3 is Q3 =199.702uC =Q4 =Q2

Now the potential difference across the C4 is given by

Q4 =C4*V4

199.702uC =(85.5uF)(V4)

V4 =199.702uC/85.5uF =2.335V

Now the potenital energy stored on the capacitor C2 is given by

PE =Q22/2C2 = (199.702uC )2/2*13.5uF =199.702*10-6)2/2*13.5*10-6 =1477.069*10-6J =1477uJ

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