In the given circuit, the potential difference across 5 {F} capacitor is [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : 48 V48\ {V}48 V

Correct Answer is : 48 V48\ {V}48 V

Test Index

Capacitor

Section: Physics

Question : 19 of 19

Marks: +1, -0

5 \mu\mathrm{F}

Solution:

According to figure,

4 \mu\mathrm{F},\ 8 \mu\mathrm{F},\ 4 \mu\mathrm{F}

are in parallel combination. So,

C'=4+8+4=16 \mu\mathrm{F}

Now,

C'

and

5 \mu\mathrm{F}

are in series combination. So, total capacitance in the circuit is,

\frac{1}{C_{\text{total }}} = \frac{1}{5} + \frac{1}{C'}

From Eq. (i),

C_{\text{total }} = \frac{16 \times 5}{16+5} = \frac{80}{21}\ \mu\mathrm{F}

Total charge in the given circuit,

q_{\text{total }} = C_{\text{total }} \cdot V

= \frac{80}{21} \times 63 = 240\ \mu\mathrm{C}

Charge remains same in series combination. So,

5 \mu\mathrm{F}

also have

240 \mu\mathrm{C}

charge.

The potential difference across

5 \mu\mathrm{F}

capacitor is,

V' = \frac{Q}{C} = \frac{240}{5}

V' = 48\ \mathrm{V}

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