When a potential difference of \(10^{3} \mathrm{~V}\) is applied between \(A\) and \(B\), a charge of \(0.75 \mathrm{mC}\) is value of \(\mathrm{C}\) is (in \(\mu \mathrm{F}\) )

In the given circuit,
\(2 \mu \mathrm{F}\) and \(2 \mu \mathrm{F}\) capacitors are in series \(\therefore\)
\(C_{S}=\frac{2 \times 2}{2+2}=1 \mu \mathrm{F}\)
So, equivalent circuit will be

Now \(C \mu \mathrm{F}\) and \(1 \mu \mathrm{F}\) are in parallel which is in series with \(1 \mu \mathrm{F}\)
\(\therefore C_{\text {eff }}=\frac{(C+1) \times 1}{(C+1)+1}=\frac{C+1}{C+2}\)
Given, \(q=0.75 \times 10^{-3} \mathrm{C}=750 \times 10^{-6} \mathrm{C}, V=10^{3} \mathrm{~V}\)
So, \(\quad C_{\text {eff }}=\frac{q}{V}\)
\(\Rightarrow \frac{C+1}{C+2}=\frac{750 \times 10^{-6}}{10^{3}}\)
\(=750 \times 10^{-3} \mathrm{~F}=0.75 \mu \mathrm{F}\)
\(\Rightarrow 0.75=\frac{C+1}{C+2}\)
\(\Rightarrow \frac{3}{4}=\frac{C+1}{C+2}\)
\(\Rightarrow 3 C+6=4 C+4 \Rightarrow C=2 \mu \mathrm{F}\)