Question

A 0.50-μF and a 1.4-μF capacitor (C_{1} and
C_{2}, respectively) are connected in series to a 14-V
battery.

a) Calculate the potential difference across each capacitor.

Express your answers using two significant figures separated by a comma.

V_{1} V_{2} =

b) Calculate the charge on each capacitor.

Express your answers using two significant figures separated by a comma.

Q_{1} Q_{2} =

c) Calculate the potential difference across each capacitor assuming the two capacitors are in parallel.

Express your answers using two significant figures separated by a comma.

V_{1} V_{2} =

d)Calculate the charge on each capacitor assuming the two capacitors are in parallel.

Express your answers using two significant figures separated by a comma.

Q_{1} Q_{2} =

Answer #1

**In series combination equivalent capacitance
is**

**C= C1×C2/ C1+ C2**

**= 0.50 ×1.40/ ( 0.50 + 1.40)
F**

**= 0.37
F**

**Now charge on each capacitor is**

**Q = CV**

**= 0.37× 14 C**

**= 5. 18
C**

**a) V1 = Q/C1**

**= 5.18/ 0.50 V**

**= 10.36 V**

**V2 = Q/C2**

**= 5.18/ 1.4 V**

**= 3. 7 V**

**b. Q1= Q2 = Q =5.18
C**

**C. In parallel combination**

**V1= V2 = 14V**

**d. Q1= C1V1**

**= 0.50 ×14 C**

**= 7.0
C**

**Q2 = C2V2**

**= 1.4 × 14 C**

**= 19.6
C**

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