Five capacitors are connected to form a Wheatstone bridge network between terminals A and B . The first branch between A and B consists of two capacitors in series having capacitances C and 3C . The second branch

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 3

Since the potential difference across the

10 \mu\text{F}

capacitor is zero, the bridge is balanced and the central capacitor can be removed from the circuit.

The equivalent capacitance of the upper branch (series combination of

C

and

3C

) is:

C_{\text{upper}} = \dfrac{C \times 3C}{C + 3C} = \dfrac{3C}{4}

The equivalent capacitance of the lower branch (series combination of

5 \mu\text{F}

and

15 \mu\text{F}

) is:

C_{\text{lower}} = \dfrac{5 \times 15}{5 + 15} = \dfrac{75}{20} = \dfrac{15}{4} \mu\text{F}

The total equivalent capacitance is the parallel combination of the upper and lower branches:

C_{\text{eq}} = C_{\text{upper}} + C_{\text{lower}}

6 = \dfrac{3C}{4} + \dfrac{15}{4}

24 = 3C + 15

3C = 9 \Rightarrow C = 3 \mu\text{F}

Answer:

3 \mu\text{F}

Five capacitors are connected to form a Wheatstone bridge network...

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