
An AC voltage is applied to a resistance $R$ and an inductor $L$ in series. If $R$ and the inductive reactance are both equal to $3\,\Omega $, the phase difference between the applied voltage and the current in the circuit is:
(A) $\dfrac{\pi }{4}$
(B) $\dfrac{\pi }{6}$
(C) $\dfrac{\pi }{2}$
Answer
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Hint The phase difference between the applied voltage and the current in the circuit is determined by using the phase difference formula, by using the given information in the formula, and taking the ${\tan ^{ - 1}}$ for the answer, the phase difference can be determined.
Useful formula
The phase difference between the applied voltage and the current in the circuit is given by,
$\tan \phi = \dfrac{{{X_L}}}{R}$
Where, $\phi $ is the phase difference between the applied voltage and the current, ${X_L}$ is the inductive reactance of the inductor and $R$ is the applied resistance of the circuit.
Complete step by step solution
Given that,
The resistance of the circuit is, $R = 3\,\Omega $,
The inductive reactance of the inductor is, ${X_L} = 3\,\Omega $.
Now,
The phase difference between the applied voltage and the current in the circuit is given by,
$\tan \phi = \dfrac{{{X_L}}}{R}\,..................\left( 1 \right)$
By substituting the inductive reactance of the inductor and the resistance in the above equation (1), then the above equation (1) is written as,
$\tan \phi = \dfrac{3}{3}$
By dividing the terms in the above equation, then the above equation is written as,
$\tan \phi = 1$
By rearranging the terms in the above equation, then the above equation is written as,
$\phi = {\tan ^{ - 1}}1$
From the trigonometry, the value of the ${\tan ^{ - 1}}\left( 1 \right) = {45^ \circ }$,
$\phi = {45^ \circ }$
Then the angle ${45^ \circ }$ is equal to the $\dfrac{\pi }{4}$, then the above equation is written as,
$\phi = \dfrac{\pi }{4}$
Thus, the above equation shows the phase difference between the applied voltage and the current in the circuit.
Hence, the option (A) is the correct answer.
NoteIn this problem we must know about the $\pi $ values for the different angles, in trigonometry there are different $\pi $ values for different angles. Here we use the angle ${45^ \circ }$ and the $\pi $ value is $\dfrac{\pi }{4}$, like that for different angles different $\pi $ values are available.
Useful formula
The phase difference between the applied voltage and the current in the circuit is given by,
$\tan \phi = \dfrac{{{X_L}}}{R}$
Where, $\phi $ is the phase difference between the applied voltage and the current, ${X_L}$ is the inductive reactance of the inductor and $R$ is the applied resistance of the circuit.
Complete step by step solution
Given that,
The resistance of the circuit is, $R = 3\,\Omega $,
The inductive reactance of the inductor is, ${X_L} = 3\,\Omega $.
Now,
The phase difference between the applied voltage and the current in the circuit is given by,
$\tan \phi = \dfrac{{{X_L}}}{R}\,..................\left( 1 \right)$
By substituting the inductive reactance of the inductor and the resistance in the above equation (1), then the above equation (1) is written as,
$\tan \phi = \dfrac{3}{3}$
By dividing the terms in the above equation, then the above equation is written as,
$\tan \phi = 1$
By rearranging the terms in the above equation, then the above equation is written as,
$\phi = {\tan ^{ - 1}}1$
From the trigonometry, the value of the ${\tan ^{ - 1}}\left( 1 \right) = {45^ \circ }$,
$\phi = {45^ \circ }$
Then the angle ${45^ \circ }$ is equal to the $\dfrac{\pi }{4}$, then the above equation is written as,
$\phi = \dfrac{\pi }{4}$
Thus, the above equation shows the phase difference between the applied voltage and the current in the circuit.
Hence, the option (A) is the correct answer.
NoteIn this problem we must know about the $\pi $ values for the different angles, in trigonometry there are different $\pi $ values for different angles. Here we use the angle ${45^ \circ }$ and the $\pi $ value is $\dfrac{\pi }{4}$, like that for different angles different $\pi $ values are available.
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