In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 KΩ with C = 2μF. The resonant frequency ω is 200 rad/s. At resonance the voltage across L is :

Question

In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 KΩ with C = 2μF. The resonant frequency ω is 200 rad/s. At resonance the voltage across L is : (A) 25×10-2 V (B) 40 V (C) 250 V (D) 4×10-3 V

Tardigrade Admin · Accepted Answer

At resonance ω L=(1/ω C) Current flowing through the circuit, I=(VR/R)=(100/1000)=0.1 A So, voltage across L is given by VL=I XL=Iω L but ω L=(1/ω C) ∴ VL=(I/ω C)=(0.1/200×2×10-6)=250 V

Q. In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 $K Ω$ with C = 2 $μ$ F. The resonant frequency $ω$ is 200 rad/s. At resonance the voltage across L is :

$25 \times 1 0^{- 2} V$

40 V

250 V

$4 \times 1 0^{- 3} V$

At resonance $ω L = \frac{1}{ω C}$
Current flowing through the circuit,
$I = \frac{V _{R}}{R} = \frac{100}{1000} = 0.1 A$
So, voltage across L is given by
$V_{L} = I X_{L} = I ω L$
$b u t ω L = \frac{1}{ω C}$
$∴ V_{L} = \frac{I}{ω C} = \frac{0.1}{200 \times 2 \times 1 0 ^{- 6}} = 250 V$

Q. In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 KΩ with C = 2μF. The resonant frequency ω is 200 rad/s. At resonance the voltage across L is :

25×10−2V