Electromagnetic Induction and Alternating Current 6 Question 1
1. A circuit connected to an $AC$ source of emf $e=e _0 \sin (100 t)$ with $t$ in seconds, gives a phase difference of $\frac{\pi}{4}$ between the emf $e$ and current $i$. Which of the following circuits will exhibit this?
(Main 2019, 8 Apr II)
(a) $R C$ circuit with $R=1 k \Omega$ and $C=1 \mu F$
(b) $R L$ circuit with $R=1 k \Omega$ and $L=1 mH$
(c) $R C$ circuit with $R=1 k \Omega$ and $C=10 \mu F$
(d) $R L$ circuit with $R=1 k \Omega$ and $L=10 mH$
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Answer:
Correct Answer: 1. (c)
Solution:
- Given, phase difference, $\varphi=\frac{\pi}{4}$
As we know, for $R-L$ or $R-C$ circuit,
$$ \begin{aligned} \Rightarrow \quad \tan \frac{\pi}{4} & =\frac{X _C \text { or } X _L}{R} \\ 1 & =\frac{X _C \text { or } X _L}{R} \Rightarrow R=X _C \text { or } X _L \end{aligned} $$
Also, given $e=e _0 \sin (100 t)$
Comparing the above equation with general equation of emf, i.e. $e=e _0 \sin \omega t$, we get
$$ \omega=100 rad / s=10^{2} rad / s $$
Now, checking option wise,
For $R-C$ circuit, with
$$ \begin{aligned} R & =1 k \Omega=10^{3} \Omega \text { and } C=1 \mu F=10^{-6} F \\ \text { So, } \quad X _C & =\frac{1}{\omega C}=\frac{1}{10^{2} \times 10^{-6}}=10^{4} \Omega \Rightarrow R \neq X _C \end{aligned} $$
For $R$ - $L$ circuit, with
$$ \begin{gathered} R=1 k \Omega=10^{3} \Omega \\ \text { and } \quad L=1 mH=10^{-3} H \\ \text { So, } X _L=\omega L=10^{2} \times 10^{-3}=10^{-1} \Omega \Rightarrow \quad R \neq X _L \end{gathered} $$
For $R$ - $C$ circuit, with
$$ \begin{array}{ll} & R=1 k \Omega=10^{3} \Omega \\ \text { and } & C=10 \mu F=10 \times 10^{-6} F=10^{-5} F \\ \text { So, } & X _C=\frac{1}{10^{2} \times 10^{-5}}=10^{3} \Omega \Rightarrow R=C \end{array} $$
For $R$ - $L$ circuit, with
$$ \begin{aligned} R & =1 k \Omega=10^{3} \Omega \\ L & =10 mH=10 \times 10^{-3} H=10^{-2} H \\ X _L & =10^{2} \times 10^{-2}=1 \Omega \Rightarrow \quad R \neq X _L \end{aligned} $$
and
Alternate Solution
Since, $\tan \frac{\pi}{4}=1=\frac{X _C \text { or } X _L}{R}$
$\therefore$ For $R-C$ circuit, we have
$$ 1=\frac{1}{C \omega R} \text { or } \omega=\frac{1}{C R} $$
Similarly, for $R-L$ circuit, we have
$$ 1=\frac{\omega L}{R} \Rightarrow \omega=\frac{R}{L} $$
It is given in the question that, $\omega=100 rad / s$
Thus, again by substituting the given values of $R, C$ or $L$ option wise in the respective Eqs. (i) and (ii), we get that only for option (c),
$$ \omega=\frac{1}{C R}=\frac{1}{10 \times 10^{-6} \times 10^{3}} \text { or } \omega=100 rad / s $$
