A circuit with an electrical load having impedance Z is connected...

A circuit with an electrical load having impedance

Z

is connected with an AC source as shown in the diagram. The source voltage varies in time as

V(t)=300 \sin (400 t) \mathrm{V}

, where

t

is time in s. List-I shows various options for the load. The possible currents

i(t)

in the circuit as a function of time are given in List-II.
A circuit with an electrical load having impedance Z is connected with an AC source as shown in the diagram. The source

A circuit with an electrical load having impedance Z is connected with an AC source as shown in the diagram. The source

Choose the option that describes the correct match between the entries in List-I to those in List-II.

	List-I		List-II
$(P)$		$(1)$
$(Q)$		$(2)$
$(R)$		$(3)$
$(S)$		$(4)$
		$(5)$

Options

A $\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$
B $\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 3$
C $\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$
D $\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5$

Correct answer

A. $\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$

Step-by-step solution

For P

i=\frac{V}{R}=10 \sin 400 t \Rightarrow(3)

For Q

\begin{aligned} & X_L=\omega L=400 \times 100 \times 10^{-3}=40 \Omega \\ & \therefore Z=50 \Omega \\ & \left.\therefore i=\frac{300}{50} \sin \left(400 t-53^{\circ}\right) \text{ [current will lag by } \tan ^{-1} \frac{X_L}{R}\right] \Rightarrow(5)\end{aligned}

For R

\begin{aligned} & \mathrm{X}_{\mathrm{C}}=\frac{10^6}{400 \times 50} \Omega=50 \Omega \text{ and } \mathrm{X}_{\mathrm{L}}=400 \times 25 \times 10^{-3}=10 \Omega \\ & \therefore \mathrm{Z}=50 \Omega \\ & \therefore \mathrm{i}=\frac{300}{50} \sin \left(400 \mathrm{t}+53^{\circ}\right) \quad\left[\text{ Current will lead by } \tan ^{-1} \frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\right] \Rightarrow(2)\end{aligned}

For

S \therefore \mathrm{i}=\frac{300}{60} \sin (400 \mathrm{t}) \quad \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \Rightarrow

Resonance

\Rightarrow(1)

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