Question

Two capacitors having capacitance value 4F, three capacitors having capacitance value 2F and 5 capacitors having capacitance value 1F are connected in parallel, calculate the equivalent capacitance.

(a) 20F

(b) 19F

(c) 18F

(d) 17F

The question was posed to me by my school teacher while I was bunking the class.

My doubt stems from Capacitors in Parallel in section Capacitance and Capacitors of Basic Electrical Engineering

Answer 1

When capacitors are connected in parallel, the equivalent capacitance is the sum of all individual capacitances.

Here, we have:

• Two capacitors with capacitance 4F each
• Three capacitors with capacitance 2F each
• Five capacitors with capacitance 1F each

The equivalent capacitance CeqC_{eq}Ceq​ is calculated as:

Ceq=(2×4)+(3×2)+(5×1)C_{eq} = (2 \times 4) + (3 \times 2) + (5 \times 1)Ceq​=(2×4)+(3×2)+(5×1)

Calculating each part:

• $2\times 4=8$
• $3\times 2=6$
• $5\times 1=5$

Now, adding these together:

Ceq=8+6+5=19 FC_{eq} = 8 + 6 + 5 = 19 \, \text{F}Ceq​=8+6+5=19F

The correct answer is:

(b) 19F