(1)

When the circuit is connected to an AC source of 12 V, it gives a current of 0.2 A.

$\text{∴ }$  $\mathrm{Impedance}, Z=\frac{12}{0.2}=60$$\Omega$

When the same circuit is connected to a DC source of 12 V, it gives a current of 0.4 A.

$\text{∴}$  $\mathrm{Resistance}, R=\frac{12}{0.4}=30$$\Omega$

As, power factor, $\mathrm{cos\varphi }=\frac{R}{Z}=\frac{30}{60}=\frac{1}{2}=\cos 60°$

$\Rightarrow \mathrm{\varphi }=60°,\text{ i.e., current lags behind the emf.}$

So, we can conclude that the circuit is a series $LR$.

(3)

Quality factor of an L-C-R circuit is given by,

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

$Q_1=\frac{1}{20}\sqrt{\frac{1.5}{35\times {10}^{-6}}}=50\times \sqrt{\frac{3}{70}}=10.35$

$Q_2=\frac{1}{25}\times \sqrt{\frac{2.5}{45\times {10}^{-6}}}=40\times \sqrt{\frac{5}{90}}=9.43$

$Q_3=\frac{1}{15}\sqrt{\frac{3.5}{30\times {10}^{-6}}}=\frac{100}{15}\sqrt{\frac{35}{3}}=22.77$

$Q_4=\frac{1}{25}\times \sqrt{\frac{1.5}{45\times {10}^{-6}}}=\frac{40}{\sqrt{30}}=7.30$

Clearly $Q_3$ is maximum of $Q_1,Q_2,Q_3$, and $Q_4$.

Hence, option (3) should be selected for better tuning of an L-C-R circuit.

(4)

Current through resistor,

$i$ = Current in the circuit

    $=\frac{V_0}{\sqrt{R^2+X_C^2}}=\frac{V_0}{\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}}$

Voltage across capacitor, $V=i{X}_C$

      $=\frac{V_0}{\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}}\times \frac{1}{\omega C}=\frac{V_0}{\sqrt{R^2{\omega }^2{C}^2+1}}$

As  $C_p<C_q$    $\therefore$  $i_p<i_q$ and $V_p>V_q$