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  3. Physics
  4. 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

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Q. In a series resonant $R-L-C$ circuit, the voltage across $R$ is $100\, V$ and the value of $R = 1000\, \Omega$. 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

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Solution:

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