First slide

Capacitance

Question

ln the circuit shown in figure the potential difference across the 4.5 $μ$ F capacitor is

Moderate

A

8 volt
B

6 volt
C

4 volt
D

8/3 volt

Solution

Let us calculate the total capacitance of the circuit. The capacitors of capacitances 3 $μ$ F and 6 $μ$ F arc in parallel, Their capacitance C' is given by $C^{'} = 3 μF + 6 μF = 9 μF$
The capacitor of capacitance C' is in series with the capacitor of capacitance 4.5 $μ$ F. Therefore, the capacitance of the system is given by $\frac{1}{C} = \frac{1}{9} + \frac{1}{4 \cdot 5} = \frac{1}{3}$
or $C = 3 μF$
The charge flowing through the circuit
$= 3 \times 12 = 36 μC$
$∴$ Potential difference a cross 4.5 $μ$ F capacitor
$= \frac{q}{C} = \frac{36}{4 \cdot 5} = 8 volt$

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