Question 469: Electrostatics & Capacitance

Question

In the circuit shown in the following figure, the potential difference cross $3 \mu \mathrm{~F}$ capacitor is
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m4us0adv/e4a717c9-f1b2-4e9a-bd4f-6246eda27d52/503c2e30-bdbb-11ef-91f0-7ffb9f85a111/file-jaoe38c1m4us0adw.png?format=png)

Accepted Answer

To find the potential difference across the $3 \, \mu \mathrm{F}$ capacitor, we start by determining the equivalent voltage and equivalent capacitance in the circuit.
First, calculate the equivalent voltage ($\mathrm{V}_{\mathrm{eq}}$):
$ \mathrm{V}_{\mathrm{eq}} = 20 - 4 = 16 \, \mathrm{V} $
Next, determine the equivalent capacitance ($\mathrm{C}_{\mathrm{eq}}$) for the parallel combination of the $3 \, \mu \mathrm{F}$ and $5 \, \mu \mathrm{F}$ capacitors:
$ \mathrm{C}_{\mathrm{eq}} = \frac{3 	imes 5}{3 + 5} = \frac{15}{8} \, \mu \mathrm{F} $
Then, calculate the total charge ($\mathrm{Q}$) using the equivalent capacitance and voltage:
$ \mathrm{Q} = \mathrm{C}_{\mathrm{eq}} 	imes \mathrm{V}_{\mathrm{eq}} = \frac{15}{8} 	imes 16 = 30 \, \mu \mathrm{C} $
Finally, find the voltage across the $3 \, \mu \mathrm{F}$ capacitor ($\mathrm{V}_3$) using the formula for charge ($\mathrm{Q}$):
$ \mathrm{V}_3 = \frac{\mathrm{Q}}{\mathrm{C}} = \frac{30}{3} = 10 \, \mathrm{V} $
The potential difference across the $3 \, \mu \mathrm{F}$ capacitor is $10 \, \mathrm{V}$.

Answer

4 V

Answer

6 V

Answer

10 V

Answer

16 V

Electrostatics & Capacitance

In the circuit shown in the following figure, the potential difference cross 3 \mu \mathrm{~F} capacitor is

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