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

Dr. Abubacker · Accepted Answer

(c) The current in the circuit
$i = \frac{V_{R}}{R}$
$= \frac{100}{1000} =$0.1 A
At resonance,
$V_{L} = V_{C} = iX_{C} = \frac{i}{\omega C}$
$= \frac{0.1}{200 	imes 2 	imes 10^{- 6}} = 250\ V$

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

Explanation

Reviewed by Dr. Abubacker

Want more practice?