If L = 25 mH, C=1mF and R = 10$\Omega$  then R.M.S current across source is

 $\begin{array}{l}{X}_L=\omega L=200\left(25\times {10}^{-3}\right)=5\Omega \text{}\\ X_C=\frac{1}{\omega C}=\frac{1}{200\times {10}^{-3}}=5\Omega \\ \therefore X_C=X_L\text{}\end{array}$ 

Current across inductor and capacitor are out of phase and of same magnitude hence only impedance is same as resistance .

$\begin{array}{l}\Rightarrow i=\frac{V}{Z}=\frac{V}{R}=\frac{50}{10}=5A\text{}\\ Now,i_{rms}=\frac{i}{\sqrt{2}}=\frac{5}{\sqrt{2}}=3.54A\end{array}$