A circuit containing inductance of $\frac{1}{6 \pi} \mathrm{H}$ and a resistance of $15 \Omega$ in series. If an $A C$ voltage of $100 \mathrm{~V}$ and $60 \mathrm{~Hz}$ is applied to above circuit, then the current in the circuit and phase difference between voltage and current respectively are

• A

  $4 \mathrm{~A}$ and $\tan ^{-1}\left(\frac{4}{5}\right)$
• B

  $5.3 \mathrm{~A}$ and $\tan ^{-1}\left(\frac{3}{4}\right)$
• Correct answer
  C

  $4 \mathrm{~A}$ and $\tan ^{-1}\left(\frac{4}{3}\right)$
• D

  5.3 $\mathrm{A}$ and $\tan ^{-1}\left(\frac{4}{3}\right)$

Solution

Circuit given is,
Inductive reactance of circuit, $$ \begin{aligned} & X_L=L \omega=L .2 \pi f \\ \Rightarrow \quad & X_L=\frac{1}{6 \pi} \times 2 \pi \times 60 \Rightarrow X_L=20 \Omega . \end{aligned} $$ Impedence of circuit, $$ \begin{aligned} Z & =\sqrt{X_L^2+R^2}=\sqrt{(20)^2+(15)^2} \\ & =\sqrt{625}=25 \Omega \end{aligned} $$ Current in circuit, $I=\frac{V}{Z}=\frac{100}{25}=4 \mathrm{~A}$ Phase difference $\phi$ between voltage and current is given by $\tan \phi=\frac{X_L}{R}=\frac{20}{15}$ $$ \Rightarrow \quad \phi=\tan ^{-1}\left(\frac{4}{3}\right) $$

Hence, the correct answer is option: 'C'.