In a series L-C-R circuit, resistance R=10 and the impedance Z=10 . The phase difference between the current and the voltage is

Correct answer: A. 0^ Impedance, aligned Z & = R^2+(X_L-X_C)^2 10 & = (10)^2+(X_L-X_C)^2 100 & =100+(X_L-X_C)^2 X_L-X_C & =0....(i) aligned Let φ is the phase difference between current and voltage aligned & & φ & =(X_L-X_C)/(R) & & φ & =(0)/(R) & & φ & =0 aligned [from Eq. (i)]

JIPMER 2010

Exam: JIPMER
Year: 2010
Subject: Physics
Chapter: Alternating Current

JIPMER 2010 · Physics · Alternating Current

In a series L-C-R circuit, resistance R=10 and the impedance Z=10 ....

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

A $0^{\circ}$
B $30^{\circ}$
C $45^{\circ}$
D $60^{\circ}$

Correct answer

A. $0^{\circ}$

Step-by-step solution

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)]

Practice Alternating Current on Quantrex Academy →

More from Alternating Current