In a series resonant R-L-C circuit, the voltage across R is 100 V and the value of R = 1000 Ω. The capacitance of the capacitor is 2 × 10-6 F; angular frequency of AC is 200 rad s-1. Then the potential difference across the inductance coil is

Question

In a series resonant R-L-C circuit, the voltage across R is 100 V and the value of R = 1000 Ω. The capacitance of the capacitor is 2 × 10-6 F; angular frequency of AC is 200 rad s-1. Then the potential difference across the inductance coil is (A) 250 V (B) 400 V (C) 100 V (D) 40 V

Tardigrade Admin · Accepted Answer

The current in the circuit i =(VR/R) =(100/1000)=0.1 A At resonance, VL =VC=i XC=(i/ω C) =(0.1/200 × 2 × 10-6) =250 V

Q. In a series resonant $R - L - C$ circuit, the voltage across $R$ is $100 V$ and the value of $R = 1000 Ω$ . The capacitance of the capacitor is $2 \times 1 0^{- 6}$ F; angular frequency of $A C$ is $200 r a d s^{- 1}$ . Then the potential difference across the inductance coil is

250 V

400 V

100 V

40 V

The current in the circuit
$i = \frac{V _{R}}{R}$
$= \frac{100}{1000} = 0.1 A$
At resonance,
$V_{L} = V_{C} = i X_{C} = \frac{i}{ω C}$
$= \frac{0.1}{200 \times 2 \times 1 0 ^{- 6}}$
$= 250 V$

Q. In a series resonant R−L−C circuit, the voltage across R is 100V and the value of R=1000Ω. The capacitance of the capacitor is 2×10−6 F; angular frequency of AC is 200rads−1. Then the potential difference across the inductance coil is