An LCR circuit contains resistance

Question:

An LCR circuit contains resistance of $110 \Omega$ and a supply of $220 \mathrm{~V}$ at $300 \mathrm{rad} / \mathrm{s}$ angular frequency. If only capacitance is removed from the

circuit, current lags behind the voltage by $45^{\circ}$. If on the other hand, only inductor is removed the current leads by $45^{\circ}$ with the applied

voltage. The rms current flowing in the circuit will be:

  1. (1) $2.5 \mathrm{~A}$

  2. (2) $2 \mathrm{~A}$

  3. (3) $1 \mathrm{~A}$

  4. (4) $1.5 \mathrm{~A}$


Correct Option: , 2

Solution:

(2)

Since $\phi$ remain same, circuit is in resonance

$\therefore \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{z}}$

$=\frac{220}{110}$

$\mathrm{I}_{\mathrm{rms}}=2 \mathrm{~A}$

Leave a comment