In above circuit if current through \(10 \Omega\) resistor is 2.5 A , value of \(R\) is ________.

C
\(10 \Omega\) and \(40 \Omega\) resistors are connected in parallel
\(\begin{array}{l} i_1 \times 10=i_2 \times 40 \\ \therefore \quad 2.5 \times 10=i_2 \times 40 \\ \left(\because i_1=2.5 A\right)\end{array}\)
\(\begin{array}{ll}\therefore & i_2=\frac{2.5 \times 10}{40} \\ \therefore & i_2=0.625 A\end{array}\)

At junction point
\(\begin{array}{l}i=i_1+i_2 \\ i=2.5+0.625 \\ i=3.125 A\end{array}\)
equivalent resistance of parallel connection of \(10 \Omega\) and \(40 \Omega\)

\(\begin{array}{l} R ^{\prime}=\frac{10 \times 40}{10+40} \\ R ^{\prime}=\frac{400}{50} \\ R ^{\prime}=8 \Omega\end{array}\)
Total resistance of circuit
\(R^{\prime \prime}=(R+8) \Omega\)
Now \(\quad V =i \times R ^{\prime \prime}\)
\(\begin{aligned} & \therefore & 50 & =3.125( R +8) \\ & \therefore & R +8 & =\frac{50}{3.125} \\ & \therefore & R +8 & =16 \\ & \therefore & R & =16-8 \\ & & R & =8 \Omega\end{aligned}\)