Question

A series LCR circuit consists of a variable capacitor connected to an inductor of inductance 50 mH,resistor of resistance 100 Ω and an AC source of angular frequency 500 rad/s. The value of capacitance so that maximum current may be drawn into the circuit is:

• A. 60 μF
• B. 50 μF
• C. 100 μF
• D. 80 μF
• E. 25 μF

Answer 1

Answer: (D)

80 μF

Answer 2

The correct option is (D): 80 μF.
To find the value of the capacitance C such that maximum current is drawn into the LCR circuit, we need to consider the condition for resonance. At resonance, the impedance of the circuit is minimized, and it occurs when the inductive reactance XL ​ and the capacitive reactance XC ​ are equal.
The angular frequency 𝜔 ω is given by:
ω =500 rad/s
The inductive reactance XL ​ is given by:
_X L_​=ωL
where 𝐿=50 mH=50×10−3 H
Substitute the values:
_X L_​=500×50×10−3
𝑋𝐿=500×0.05 XL ​=500×0.05
𝑋𝐿=25 Ω
At resonance, the capacitive reactance XC ​ is equal to the inductive reactance XL ​:
_X C_​=_X L_​
$\frac{1}{ωc}$​=25
Now, solve for C :
$C=\frac{1}{ωXC​}​$
$𝐶=\frac{1}{500×25}​$
$𝐶=\frac{1}{12500}$
𝐶=8×10−5
C =80 μ F
Thus, the capacitance required for maximum current to be drawn into the circuit is 80 μ F