List-I contains different AC circuits with rms voltage of ac sources as shown. List-II  contains the corresponding rms currents supplied by the ac voltage sources. Then match  the options of List-I currently with the options List-II.

	List-I		List-II
P)		1)	${\text{I}}_{rms}=25A$
Q)		2)	${\text{I}}_{rms}=5A$
R)		3)	${\text{I}}_{rms}=10\sqrt{5}A$
S)		4)	${\text{I}}_{rms}=20A$
		5)	${\text{I}}_{rms}=30A$

The correct options is:

  (P)  $z=R+\left(X_L-X_C\right)i=3+(8-4)i=3+4i$
 $$\begin{gathered} \left|z\right|=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}=5\Omega \\ {l}_{rms}=\frac{{\epsilon }_{rms}}{\left|z\right|}=\frac{100}{5} \end{gathered}$$
 
(Q) $\frac{1}{z}=\frac{1}{X_{L}i}+\frac{1}{-X_{c}i}+\frac{1}{R}=\frac{1}{6i}-\frac{1}{3i}+\frac{1}{8}=\left(\frac{1}{3}-\frac{1}{6}\right)i+\frac{1}{8}=\frac{1}{8}+\frac{1}{6}i$
 $$\begin{gathered} \Rightarrow z=\frac{48}{\left(6+8i\right)} \\ \Rightarrow z=\frac{48(6-8i)}{100} \\ \Rightarrow \left|z\right|=\frac{48}{100}\times 10 \\ \Rightarrow \left|z\right|=4.8\Omega \\ {l}_{rms}=\frac{{\epsilon }_{rms}}{\left|z\right|}=\frac{120}{4.8}=25A \end{gathered}$$
 (R)  $\frac{1}{z^{\text{'}}}=\frac{1}{X_{L}i}+\frac{1}{-X_{c}i}=\frac{1}{6i}-\frac{1}{3i}=i\left(\frac{1}{3}-\frac{1}{6}\right)=\frac{i}{6}$
 $$\begin{gathered} z^{\text{'}}=\frac{6}{i}=-6i \\ z=R+z^{\text{'}}=8-6i \\ \left|z\right|=\sqrt{{\left(8\right)}^2+{\left(6\right)}^2}=10\Omega \\ {l}_{rms}=\frac{{\epsilon }_{rms}}{\left|z\right|}=\frac{50}{10}=5A \end{gathered}$$
 (S)  $z_1=R_1+X_{L}i=8+6i\Rightarrow I_{1\left(rms\right)}=\frac{{\epsilon }_{rms}}{\left|z_1\right|}=\frac{100}{10}=10A$
 $$\begin{gathered} z_2=R_2+X_{C}i=3+4i\Rightarrow I_{2\left(rms\right)}=\frac{{\epsilon }_{rms}}{\left|z_2\right|}=\frac{100}{5}=20A \\ \therefore I_{rms}=\sqrt{I_{1\left(rms\right)}^2+I_{2\left(rms\right)}^2}=\sqrt{{\left(10\right)}^2+{\left(20\right)}^2}\sqrt{500}=10\sqrt{5}A \end{gathered}$$