Question

An AC circuit consists of a series combination of an inductance $L=1mH$, a resistance $R=1\Omega$ and a capacitance $C$. It is observed that the current leads the voltage by $45°$. Find the value of capacitance '$C$' if the angular frequency of applied AC is $300rad/s$.

• A. $5.6mF$
• B. $3.92mF$
• C. $2.56mF$
• D. $5.2mF$

Answer 1

The correct option is C

$2.56mF$

Step 1: Given data:

Inductance, $L=1mH$

$=1\times {10}^{-3}H$

Capacitance $=C$

Resistance $R=1\Omega$

Angular frequency $\omega =300rad/s$

Phase difference $\theta =45°$.

Step 3: Formula used:

$\tan \theta =\frac{\left|X_C-X_L\right|}{R}$

where, $X_c=$capacitive reactance

$=\frac{1}{\omega C}$

$X_L=$inductive reactance

$=\omega L$

$\theta =$ phase difference

$R=$resistance

Step 3: Calculating the Reactance:

Capacitive reactance, $X_C=\frac{1}{300C}\Omega$

Inductive reactance, $X_L=300\times 0.001$

$=0.3\Omega$

Step 4: Calculating the capacitance:

$\tan 45°=\frac{\left|{\displaystyle \frac{1}{300C}}-0.3\right|}{1}$

$\frac{1}{300C}-0.3=1$

$\frac{1}{300C}=1.3$

$C=\frac{1}{300\times 1.3}$

$=0.00256F$

$=2.56mF$

Hence, option C is the correct answer.