Search any question & find its solution
Question:
Answered & Verified by Expert
In a series $L-C-R$ circuit, resistance $R=10 \Omega$ and the impedance $Z=10 \Omega$. The phase difference between the current and the voltage is
Options:
Solution:
1877 Upvotes
Verified Answer
The correct answer is:
$0^{\circ}$
Impedance,
$\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
\therefore \quad 10 & =\sqrt{(10)^2+\left(X_L-X_C\right)^2} \\
\Rightarrow \quad 100 & =100+\left(X_L-X_C\right)^2 \\
\Rightarrow X_L-X_C & =0....(i)
\end{aligned}$
Let $\phi$ is the phase difference between current and voltage
$\begin{aligned}
\therefore & & \tan \phi & =\frac{X_L-X_C}{R} \\
\therefore & & \tan \phi & =\frac{0}{R} \\
& \Rightarrow & \phi & =0
\end{aligned}$
[from Eq. (i)]
$\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
\therefore \quad 10 & =\sqrt{(10)^2+\left(X_L-X_C\right)^2} \\
\Rightarrow \quad 100 & =100+\left(X_L-X_C\right)^2 \\
\Rightarrow X_L-X_C & =0....(i)
\end{aligned}$
Let $\phi$ is the phase difference between current and voltage
$\begin{aligned}
\therefore & & \tan \phi & =\frac{X_L-X_C}{R} \\
\therefore & & \tan \phi & =\frac{0}{R} \\
& \Rightarrow & \phi & =0
\end{aligned}$
[from Eq. (i)]
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.