Q.42. The figure given below shows an LCR series circuit with two switches and . When switch is cl

Q: What is the correct answer for this exam Physics Q42?

The correct answer is B. Case 1: $S_1$ closed, $S_2$ open. The circuit consists of $R$, $C$, and $L_1$. The phase difference is $30^\circ$. $\tan \phi_1 = \frac{|X_{L1} - X_C|}{R} = \tan 30^\circ = \frac{1}{\sqrt{3}}$. So, $|\omega L_1 - \frac{1}{\omega C}| = \frac{R}{\sqrt{3}}$ (Equation 1). Case 2: $S_2$ closed, $S_1$ open. The circuit consists of $R$, $C$, and $L_2$. The phase difference is $60^\circ$. $\tan \phi_2 = \

Q: Q.42. The figure given below shows an LCR series circuit with two switches and . When switch is closed keeping open, the phase difference between the current and source voltage is and phase…

The correct answer is B. Case 1: $S_1$ closed, $S_2$ open. The circuit consists of $R$, $C$, and $L_1$. The phase difference is $30^\circ$. $\tan \phi_1 = \frac{|X_{L1} - X_C|}{R} = \tan 30^\circ = \frac{1}{\sqrt{3}}$. So, $|\omega L_1 - \frac{1}{\omega C}| = \frac{R}{\sqrt{3}}$ (Equation 1). Case 2: $S_2$ closed, $S_1$ open. The circuit consists of $R$, $C$, and $L_2$. The phase difference is $60^\circ$. $\tan \phi_2 = \

Q: What are the options for this this exam Physics question?

A. 1/3 · B. 2/9 · C. 1/5 · D. 4/5

Question

Q.42. The figure given below shows an LCR series circuit with two switches  and . When switch  is closed keeping  open, the phase difference between the current and source voltage is  and phase difference is  when  is closed keeping  open. The value of  is __________ .

FirstInTest · Accepted Answer

Case 1: $S_1$ closed, $S_2$ open. The circuit consists of $R$, $C$, and $L_1$. The phase difference is $30^\circ$.
$	an \phi_1 = \frac{|X_{L1} - X_C|}{R} = 	an 30^\circ = \frac{1}{\sqrt{3}}$.
So, $|\omega L_1 - \frac{1}{\omega C}| = \frac{R}{\sqrt{3}}$ (Equation 1).
Case 2: $S_2$ closed, $S_1$ open. The circuit consists of $R$, $C$, and $L_2$. The phase difference is $60^\circ$.
$	an \phi_2 = \frac{|X_{L2} - X_C|}{R} = 	an 60^\circ = \sqrt{3}$.
So, $|\omega L_2 - \frac{1}{\omega C}| = R\sqrt{3}$ (Equation 2).
From the solution, it seems that $X_{L1} > X_C$ and $X_{L2} > X_C$ are assumed, or the absolute value is handled by the context of the problem (e.g., inductive circuit).
From (1): $\omega L_1 - \frac{1}{\omega C} = \frac{R}{\sqrt{3}}$ (assuming inductive circuit)
From (2): $\omega L_2 - \frac{1}{\omega C} = R\sqrt{3}$ (assuming inductive circuit)
Subtracting (1) from (2) (or vice versa, depending on which is larger):
$(\omega L_2 - \frac{1}{\omega C}) - (\omega L_1 - \frac{1}{\omega C}) = R\sqrt{3} - \frac{R}{\sqrt{3}}$.
$\omega L_2 - \omega L_1 = R (\sqrt{3} - \frac{1}{\sqrt{3}}) = R (\frac{3-1}{\sqrt{3}}) = \frac{2R}{\sqrt{3}}$.
We need to find $(3L_1 - L_2)$. This implies we need to find $L_1$ and $L_2$ individually or a relation.
The given source voltage is $V = V_0 \sin(300t)$, so $\omega = 300 \mathrm{~rad/s}$.
$C = 100 \mu \mathrm{F} = 100 	imes 10^{-6} \mathrm{~F} = 10^{-4} \mathrm{~F}$.
$\frac{1}{\omega C} = \frac{1}{300 	imes 10^{-4}} = \frac{1}{3 	imes 10^{-2}} = \frac{100}{3} \Omega$.
From (1): $300 L_1 - \frac{100}{3} = \frac{R}{\sqrt{3}}$.
From (2): $300 L_2 - \frac{100}{3} = R\sqrt{3}$.
Divide (2) by (1): $\frac{300 L_2 - 100/3}{300 L_1 - 100/3} = \frac{R\sqrt{3}}{R/\sqrt{3}} = 3$.
$300 L_2 - \frac{100}{3} = 3 (300 L_1 - \frac{100}{3})$.
$300 L_2 - \frac{100}{3} = 900 L_1 - 100$.
$300 L_2 = 900 L_1 - 100 + \frac{100}{3} = 900 L_1 - \frac{300 - 100}{3} = 900 L_1 - \frac{200}{3}$.
Divide by 100: $3 L_2 = 9 L_1 - \frac{2}{3}$.
We need to find $(3L_1 - L_2)$.
From $3 L_2 = 9 L_1 - \frac{2}{3}$, we have $L_2 = 3 L_1 - \frac{2}{9}$.
So, $3L_1 - L_2 = 3L_1 - (3L_1 - \frac{2}{9}) = 3L_1 - 3L_1 + \frac{2}{9} = \frac{2}{9} \mathrm{~H}$.

Answer

1/3

Answer

2/9

Answer

1/5

Answer

4/5

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