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,
[tex]C_0=C_2+C_3[/tex]
On substituting the known values,
[tex]\begin{gathered} C_0=20\text{ }\mu F+40\text{ }\mu F \\ =60\text{ }\mu\text{F} \end{gathered}[/tex]
The voltage across the equivalent resistance and thus across each of the capacitors connected in parallel is given by,
[tex]V_0_{}=\frac{C_1}{C_1+C_0}\times V[/tex]
On substituting the known values,
[tex]\begin{gathered} V_0=\frac{150\times10^{-6}}{150\times10^{-6}+60\times10^{-6}}\times400 \\ =285.7\text{ V} \end{gathered}[/tex]
The charge across the 40 μF capacitor is given by,
[tex]Q=C_3V_0_{}[/tex]
On substituting the known values,
[tex]\begin{gathered} Q=40\times10^{-6}\times285.7 \\ =0.0114\text{ C} \end{gathered}[/tex]
Thus the charge across the 40 μF capacitor is 0.0114 C
Therefore the correct answer is option A.
