Question

(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source.
(b) What is the phase difference between the voltages across inductor and the capacitor at resonance in the LCR circuit ?
(c) When an inductor is connected to a 200 V dc voltage, a current of 1A flows through it. When the same inductor is connected to a 200 V, 50 Hz ac source, only 0.5 A current flows. Explain, why ? Also, calculate the self inductance of the inductor.

Answer 1

(a) Expression for its impedance:

From question, a series LCR circuit connected across an ac source of variable frequency.

Voltage of ac source:

$\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$

Current of ac source:

$\mathrm{I}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$

On applying Ohm's law, the voltage drop across R is given by the formula,

$\mathrm{V}_{\mathrm{R}}=\mathrm{I}_{\mathrm{m}} \mathrm{R}$

The voltage is in same phase with current.

The voltage across inductor is given by the formula,

$V_{L}=I_{m} X_{L}$

Where,

$X_{L}$ = Inductance = $\omega C$

Now, the voltage leads the current by π/2.

The voltage across capacitor is given by the formula,

$\mathrm{V}_{\mathrm{c}}=\mathrm{I}_{\mathrm{m}} \mathrm{X}_{\mathrm{c}}$

$X_{c}$ = Capacitive reactance = $X_{c}=1 / \omega c$

Now, the voltage lags the current by π/2.

The resultant voltage is (refer the graph given below):

$V_{\mathrm{m}}=\sqrt{V_{\mathrm{R}}^{2}+\left(V_{\mathrm{L}}-V_{\mathrm{C}}\right)^{2}}$

$V_m=\sqrt{\left(I_{\mathrm{m}} R\right)^{2}+\left(I_{\mathrm{m}} X_{\mathrm{L}}-I_{\mathrm{m}} X_{\mathrm{C}}\right)^{2}}$

$V_m=I_{\mathrm{m}} \sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}$

$\frac{V_{\mathrm{m}}}{I_{\mathrm{m}}}=\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}$

Now, the resistance is ratio of voltage to current.

So, impedance is $\mathrm{V}_{\mathrm{m}} / \mathrm{I}_{\mathrm{m}}=\mathrm{z}$

$z=\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}$

Thus, the expression for its impedance is obtained.

(b) Phase difference between inductor and the capacitor:

From question, inductor and the capacitor are at resonance. Thus, the phase angle of circuit is zero.

$\tan \phi=\frac{X_{\mathrm{L}}-X_{\mathrm{C}}}{R}=0$

$X_{\mathrm{L}}-X_{\mathrm{C}}=0$

Therefore,

$X_{\mathrm{L}}=X_{\mathrm{C}}$

Thus, the impendence is:

$z=\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}$

$\Rightarrow z=\sqrt{R^{2}} = R$

From this, we can understand that the voltage across inductance and capacitance is same. The voltage drop across impedance is equal to impedance across resistance.

Thus, the phase difference between the voltages across inductor and the capacitor at resonance in the LCR circuit is 180°.

(c) Self inductance of the inductor:

From question, the inductor is connected to dc voltage, so it acts as resistor.

The resistance across the coil is given by the formula,

$R=V / I = 200/1 = 200 \ \Omega$

From question, same inductor is connected to a 200 V, 50 Hz ac source.

Angular frequency is:

$\omega=2 \pi v=2 \pi \times 50$

Net impedance is:

$Z=\sqrt{X_{L}^{2}+R^{2}}$

On applying Ohm's law, we get,

$Z=\frac{V}{I}$

$Z=\frac{200}{0.5}$

$\therefore Z=400 \ \Omega$

Now, the impedance is:

$400=\sqrt{X_{L}^{2}+R^{2}}$

$400=\sqrt{\left(W_{L}\right)^{2}+R^{2}}$

On substituting the values, we get,

$400=\sqrt{\left(2 \pi \times 50 \times X_{L}\right)^{2}+(200)^{2}}$

On squaring both sides, we get,

$(400)^{2}=(2 \pi \times 50 \times L)^{2}+(200)^{2}$

$160000=(100 \pi L)^{2}+40000$

$120000=10000 \pi^{2} L^{2}$

$L^{2}=\frac{120000}{10000 \pi^{2}}$

Thus, the self-inductance of inductor is:

$L=\left(\frac{\sqrt{12}}{\pi}\right) \mathrm{H}$