Question 52: AC fundamentals

Question

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.

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

List–I
List–II

(P) 
(1)

(Q) 
(2)

(R) 
(3)

(S) 
(4)

(5)

Accepted Answer

$$ \begin{aligned} & \Rightarrow i_0=\frac{v_0}{R}=\frac{300}{30}=10 \mathrm{~A} \ & 	ext { and } \phi=0 \ & \begin{array}{l}  So (P) ightarrow 3 \end{array} \end{aligned} $$
$$ \begin{aligned} z & =\sqrt{R^2+(\omega L)^2} \quad\left(	ext { as } x_L=\omega Light) \ & =\sqrt{(30)^2+\left(400  	imes 100 	imes 10^{-3}ight)^2} \ & =\sqrt{900+1600}=\sqrt{2500} \ z=50 \Omega & \end{aligned} $$
$$ \begin{aligned} 	ext { So } & i_0=\frac{V_0}{2}=\frac{300}{50}=6 \mathrm{~A} \ \phi & =	an ^{-1}\left(\frac{x_L}{R}ight)=	an ^{-1}\left(\frac{\omega L}{R}ight) \ & =	an ^{-1}\left(\frac{400 	imes 106 	imes 10^{-3}}{36}ight)=	an ^{-1}\left(\frac{4}{3}ight) \ \Rightarrow \quad \phi & =53^{\circ}(l \mathrm{ag}) \ & 	herefore Q ightarrow 5 \end{aligned} $$
$$ \begin{aligned} z & =\sqrt{R^2+\left(x_C-x_Light)^2} \ & =\sqrt{R^2+\left(\frac{1}{\omega c}-\omega Light)^2} \ & =\sqrt{(30)^2+\left(\frac{1}{400 	imes 50 	imes 10^{-6}}-400 	imes 25 	imes 10^{-3}ight)^2} \ z & =\sqrt{900+\left(\frac{1}{20 	imes 10^{-3}}-10ight)^2} \end{aligned} $$
$$ \begin{aligned} &\begin{array}{r} z=\sqrt{900+(50-10)^2}=\sqrt{900+1600} \ z=\sqrt{2500}=50 \Omega \ i_0=\frac{v_0}{2}=\frac{300 }{50}=6 \mathrm{~A} \ \phi=	an ^{-1}\left[\frac{x_L-x_C}{R}ight]=	an ^{-1}\left[\frac{10-50}{30}ight] \ =	an ^{-1}\left|\frac{40 }{30}ight|=53^{\circ} 	ext { (Lead) } \end{array}\ &(R) ightarrow 2 \end{aligned} $$
$$ \begin{aligned} z & =\sqrt{(60)^2+\left(\frac{1}{400 	imes 50 	imes 10^{-6}}-400 	imes 125 	imes 10^{-3}ight)^2} \ & =\sqrt{(60)^2+(50-50)^2} \ z & =60 \Omega \ & i_0=\frac{300}{60}=5 \mathrm{~A} \ \phi & =	an ^{-1}\left|\frac{x_L-x_C}{R}ight|=	an ^{-1}\left|\frac{0}{60}ight|=0^{\circ} \end{aligned} $$
So $(s) ightarrow 1$
Hence, option (A) is correct.

Answer

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

Answer

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

Answer

$\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$

Answer

$\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5$

AC fundamentals

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