# Basic Electrical Engineering Questions and Answers – Capacitors in Series

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This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Capacitors in Series”.

1. What is the total capacitance when two capacitors C1 and C2 are connected in series?
a) (C1+C2)/C1C2
b) 1/C1+1/C2
c) C1C2/(C1+C2)
d) C1+C2

Explanation: When capacitors are connected in series, the equivalent capacitance is:
1/Ctotal=1/C1+1/C2, therefore Ctotal = C1C2/(C1+C2).

2. N capacitors having capacitance C are connected in series, calculate the equivalent capacitance.
a) C/N
b) C
c) CN
d) N/C

Explanation: When capacitors are connected in series, the equivalent capacitance is:
1/Ctotal = 1/C+1/C+1/C+……..N times.
1/Ctotal=N/C.
Ctotal=C/N.

3. When capacitors are connected in series, the equivalent capacitance is ___________ each individual capacitance.
a) Greater than
b) Less then
c) Equal to
d) Insufficient data provided

Explanation: When capacitors are connected in series, the equivalent capacitance is:
1/Ctotal=1/C1+1/C2. Since we find the reciprocals of the sum of the reciprocals, the equivalent capacitance is less than the individual capacitance values.

4. What is the equivalent capacitance? a) 1.5F
b) 0.667F
c) 2.45F
d) 2.75F

Explanation: When capacitors are connected in series,
1/Ctotal = 1/C1+1/C2 = 1/2+1 = 3/2
Ctotal = 2/3 = 0.667F.

5. When capacitors are connected in series ___________ remains the same.
a) Voltage across each capacitor
b) Charge
c) Capacitance
d) Resistance

Explanation: When capacitors are connected in series, the charge remains the same because the same amount of current flow exists in each capacitor.

6. When capacitors are connected in series _______________ Varies
a) Voltage across each capacitor
b) Charge
c) Capacitance
d) Resistance

Explanation: When capacitors are connected in series, the voltage varies because the voltage drop across each capacitor is different.

7. Four 10F capacitors are connected in series, calculate the equivalent capacitance.
a) 1.5F
b) 2.5F
c) 3.5F
d) 0.5F

Explanation: When capacitors are connected in series,
1/Ctotal=1/C1+1/C2+1/C3+1/C4=1/10+1/10+1/10+1/10=4/10F.
Ctotal=10/4=2.5F.

8. Calculate the charge in the circuit. a) 66.67C
b) 20.34C
c) 25.45C
d) 30.45C

Explanation: When capacitors are connected in series, the equivalent capacitance is:
1/Ctotal=1/C1+1/C2 = 1/2+1=3/2
Ctotal=2/3 F
Q=CV=(2/3)*100 = 200/3 C=66.67C.

9. Calculate the voltage across the 1F capacitor. a) 33.33V
b) 66.67V
c) 56.56V
d) 23.43V

Explanation: When capacitors are connected in series,
1/Ctotal=1/C1+1/C2 = 1/2+1=3/2
Q = CV = (2/3)*100 = 66.67C.
V across the 1F capacitor = 66.67/1 = 66.67V.

10. Calculate the voltage across the 2F capacitor. a) 33.33V
b) 66.67V
c) 56.56V
d) 23.43V

Explanation: When capacitors are connected in series,
1/Ctotal=1/C1+1/C2 = 1/2+1 = 3/2
Q = CV = (2/3)*100 = 66.67C.
V across the 2F capacitor = 66.67/2 = 33.33V.

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