A capacitor of 10\,{F} charged upto 250\,{V} is connected in parallel with another capacitor of 5\,{F} charged upto 100\,{V}. The common potential is

Correct Answer is : 200 V200\,{V}200V

Correct Answer is : 200 V200\,{V}200V

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CGPET 2021 Solved Paper

Section: Physics

Question : 7 of 150

Marks: +1, -0

A capacitor of

10\,\mu\mathrm{F}

250\,\mathrm{V}

5\,\mu\mathrm{F}

100\,\mathrm{V}

Solution:

Given, capacitance of first capacitor,

C_1 = 10\,\mu\mathrm{F}

Potential difference across first capacitor,

V_1 = 250\,\mathrm{V}

Capacitance of second capacitor,

C_2 = 5\,\mu\mathrm{F}

Potential difference across second capacitor,

V_2 = 100\,\mathrm{V}

Since,

C_1 \text{ and } C_2

are parallel,

∴

C_{\text{eff}} = C_1 + C_2

and total charge,

Q = Q_1 + Q_2

Here,

Q_1 = V_1C_1 \text{ and } Q_2 = V_2C_2

Now, common potential,

V = \frac{Q}{C_{\text{eff}}} = \frac{Q_1+Q_2}{C_1+C_2}

V = \frac{V_1C_1 + V_2C_2}{C_1+C_2}

= \frac{250 \times 10 \times 10^{-6} + 100 \times 5 \times 10^{-6}}{10 \times 10^{-6} + 5 \times 10^{-6}}

= \frac{2500+500}{15} = \frac{3000}{15} = 200\,\mathrm{V}

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