Question

For the given diagram, determine the charge stored on the 40 μF capacitor.0.0114 C0.0123 C0.0136 C0.0110 C

Answer 1

Given,

The supply voltage, V=400 V

The capacitance of capacitors;

C₁=150 μF

C₂=20 μF

C₃=40 μF

The equivalent capacitance of the capacitors connected in the parallel is given by,

`C_0=C_2+C_3`

On substituting the known values,

`\begin{gathered} C_0=20\text{ }\mu F+40\text{ }\mu F \\ =60\text{ }\mu\text{F} \end{gathered}`

The voltage across the equivalent resistance and thus across each of the capacitors connected in parallel is given by,

`V_0_{}=(C_1)/(C_1+C_0)* V`

On substituting the known values,

`\begin{gathered} V_0=(150*10^(-6))/(150*10^(-6)+60*10^(-6))*400 \\ =285.7\text{ V} \end{gathered}`

The charge across the 40 μF capacitor is given by,

`Q=C_3V_0_{}`

On substituting the known values,

`\begin{gathered} Q=40*10^(-6)*285.7 \\ =0.0114\text{ C} \end{gathered}`

Thus the charge across the 40 μF capacitor is 0.0114 C

Therefore the correct answer is option A.