Question 1. The circuit at right contains 3 capacitors. (a) What is the charge,

Question

Question 1. The circuit at right contains 3 capacitors. (a) What is the charge, Q1, on capacitor C1? (b) What is the charge, Q2, on capacitor C2? (c) Now suppose C3 is modified by filling it with a dielectric with constant K> 1. How does the charge, Q2, on C2 change? i. Q2 decreases because the capacitance of C2 decreases relative to that of C3. ii. Q2 stays the same because the charge of capacitors in series is the same. iii. Q2 increases because the capacitance of C3 increases. (d) How does the charge, Q1, on C1 change when C3 is modified by filling it with a dielectric with constant K> 1? i. Q1 decreases because the capacitance of C1 decreases relative to that of C3. ii. Q1 stays the same because the voltage across C1 is the same as it was originally. iii. Q1 increases because the effective capacitance of C2 C3 part of the circuit increases.

Explanation / Answer

a)

In the given circuit C1 is in parallel to series combination of C2 and C3

In parallel Voltage across Capacitors will be same i.e

V1=V23=V

Charge on C1 is

Q1=C1V1 =C1V

b)

Voltage across C2 is

V2=V*(C2/C2+C3)

Charge on C2 is

Q2=C2V2 =C2*[V*(C2/C2+C3)]

Q2=[C22/(C2+C3)]*V

c)

1 is The correct answer.

With Dielectric inserted K>1 ,the capacitance Increses by a factor K

C2'=KC3

So charge on C2 is

Q2=Q2=[C22/(C2+KC3)]*V

So from above equation charge on Capacitor C2 decreases

d)

2 is the correct answer ,as voltage remains same across C1

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