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capacitance

In the given circuit, a charge of $+80 \mu C$ is given to...

2 months ago 1

In the given circuit, a charge of

+ 80 μ C

$+80 \mu C$ is given to the upper plate of the

4 μ F

$4 \mu F$ capacitor. Then in the steady state, the charge on the upper plate of the

3 μ F

$3 \mu F$ capacitor is

+ 32 μ C

$+32 \mu \mathrm{C}$

+ 40 μ C

$+40 \mu \mathrm{C}$

+ 48 μ C

$+48 \mu \mathrm{C}$

+ 80 μ C

$+80 \mu \mathrm{C}$

Solution:

The total charge on plate

A

$A$ will be

80 μ C

$80 \mu \mathrm{C}$ .

2 μ F

$2 \mu F$ and

3 μ F

$3 \mu F$ capacitors are in parallel. Therefore,

C_{e q} = 2 + 3 = 5 H F

$C_{e q}=2+3=5 \mathrm{HF}$

Charge on capacitor of

3 μ F

$3 \mu \mathrm{F}$ capacitance

q = \frac{3}{5} \times 80 = 48 μ C

$q=\frac{3}{5} \times 80=48 \mu C$