In a series resonant L-C-R circuit, the voltage across R is 100 V and R = 1 k Ω with C = 2 μ F. The resonant frequency ω is 200 rad s−1. At resonance the voltage across L is

Question

In a series resonant L-C-R circuit, the voltage across R is 100 V and R = 1 k Ω with C = 2 μ F. The resonant frequency ω is 200 rad s−1. At resonance the voltage across L is (A) 2.5 × 10−2 V (B) 4 × 10−8 V (C) 250 V (D) 2.5 × 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=l ω L but ω L =(1/ω C) ∴ VL =(1/ω C) =(0.1/200 × 2 × 10-6) =250 V

Q. In a series resonant $L - C - R$ circuit, the voltage across $R$ is $100 V$ and $R = 1 k Ω$ with $C = 2 μ F$ . The resonant frequency $ω$ is $200 r a d s^{- 1}$ . At resonance the voltage across $L$ is

$2.5 \times 1 0^{- 2} V$

$4 \times 1 0^{- 8} V$

$250 V$

$2.5 \times 1 0^{- 3} V$

Q. In a series resonant L−C−R circuit, the voltage across R is 100V and R=1kΩ with C=2μF. The resonant frequency ω is 200rads−1. At resonance the voltage across L is

2.5×10−2V

4×10−8V

250V

2.5×10−3V

At resonance, ωL=ωC1​ Current flowing through the circuit, I=RVR​​=1000100​ =0.1A So, voltage across L is given by VL​=IXL​=lωL but ωL=ωC1​ ∴VL​=ωC1​ =200×2×10−60.1​ =250V

Q. In a series resonant $L - C - R$ circuit, the voltage across $R$ is $100 V$ and $R = 1 k Ω$ with $C = 2 μ F$ . The resonant frequency $ω$ is $200 r a d s^{- 1}$ . At resonance the voltage across $L$ is

$2.5 \times 1 0^{- 2} V$

$4 \times 1 0^{- 8} V$

$250 V$

$2.5 \times 1 0^{- 3} V$