Alternating Current

For a series $L$-$C$-$R$ circuit:

$ Z = \sqrt{R^2 + (X_L - X_C)^2} $

Step 1: What does “purely resistive” mean?

If the circuit is purely resistive, it means only the resistor affects the circuit. So, the total impedance $Z$ should equal the resistance $R$.

Step 2: Set impedance equal to resistance:

$ Z = R $

So, $ R = \sqrt{R^2 + (X_L - X_C)^2} $

Step 3: Solve for $X_L$ and $X_C$:

$ R^2 = R^2 + (X_L - X_C)^2 $

Subtract $R^2$ from both sides: $ 0 = (X_L - X_C)^2 $

This means: $ X_L = X_C $

Conclusion:

If the circuit is purely resistive, the inductive reactance $X_L$ must be equal to the capacitive reactance $X_C$.

$ \begin{aligned} &\\ &\text { } \begin{aligned} \tan \phi & =\frac{X_C}{R}=\frac{1}{\omega C R} \\ C & =\frac{1}{\omega R \tan \phi} \\ & =\frac{1}{200 \times 100 \sqrt{3} \tan 30^{\circ}} \\ & =\frac{1}{2 \times 10^4}=50 \times 10^{-6} \mathrm{~F}=50 \mu \mathrm{~F} \end{aligned} \end{aligned} $

When iron rod is inserted into interior of the inductor, then its self inductance increases.

Since, $X_L=\omega L$

i.e., $X_L \propto L$

Thus, due to increase in the value of $L$, $X_L$ also increases.

Hence, current decreases.

Thus, the glow of the bulb decreases.

• In an L-R series circuit,

• Voltage across resistor: $ V_R = 240 \, \text{V} $

• Voltage across inductor: $ V_L = 180 \, \text{V} $

We are asked to find the supply voltage $ V $.

Step 1: Relation between voltages in series L-R circuit

In a series L-R circuit, the resistor voltage $ V_R $ and the inductor voltage $ V_L $ are 90° out of phase (since $ V_R $ is in phase with current, and $ V_L $ leads the current by 90°).

Hence, the total voltage is obtained by the vector sum:

$ V = \sqrt{V_R^2 + V_L^2} $

Step 2: Substitute given values

$ V = \sqrt{(240)^2 + (180)^2} $

$ V = \sqrt{57600 + 32400} = \sqrt{90000} = 300 \, \text{V} $

✅ Final Answer:

300 V

Option A

$ \begin{aligned} &\text { } V=V_0 \sin \omega t\\ &\begin{aligned} & \Rightarrow \frac{V_0}{2}=V_0 \sin \omega t \\ & \Rightarrow \frac{1}{2}=\sin \omega t \Rightarrow \omega t=\frac{\pi}{6} \\ & \Rightarrow 2 \pi f t=\frac{\pi}{6} \Rightarrow 2 \pi \times 50 \times t=\frac{\pi}{6} \\ & \Rightarrow \quad t=\frac{1}{600} \mathrm{~s} \end{aligned} \end{aligned} $

Given,

The ratio of the capacitive reactance $ X_{C} $ to the resistance $ R $ is $ 4: 3 $.

Let, $ X_{C}=4 k $

$ R=3 k $

Then, the impedance $ Z $ of the series $ R-C $ circit is given by

$ \begin{array}{l} Z=\sqrt{R^{2}+X_{C}^{2}} \\\\ Z=\sqrt{(3 k)^{2}+(4 k)^{2}} \\\\ Z=5 k \end{array} $

We know that, the power factor is the cosine of the phase angle $ \phi $ between the total impedance and the resistive part of the impedance.

$ \begin{array}{l} \cos \phi=\frac{R}{Z} \\\\ \cos \phi=\frac{3 k}{5 k} \\\\ \cos (\phi)=0.6 \end{array} $

Given,

Inductive reactance, $ X_{L}=R $

Capacitive reactance, $ X_{C}=2 R $

Resistor $ =R $

Power factor, $ \cos \phi $ is given by

$ \begin{array}{l} \cos \phi=\frac{R}{Z} \\\\ \cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}} \\\\ \cos \phi=\frac{R}{\sqrt{R^{2}+(R-2 R)^{2}}} \\\\ \cos \phi=\frac{R}{\sqrt{2 R^{2}}}=\frac{1}{\sqrt{2}} \end{array} $

Given, $ V_{L}=6 \mathrm{~V} $

$ V_{\text {RMS }}=10 \mathrm{~V} $

As the circuit consists only $ L $ and $ R $

$ V_{\mathrm{RMS}}=\sqrt{V_{R}^{2}+V_{L}^{2}} $

Squaring both sides

$ \begin{aligned} V_{\mathrm{RMS}}^{2} & =V_{R}^{2}+V_{L}^{2} \\\\ V_{R}^{2} & =V_{\mathrm{RMS}}^{2}-V_{L}^{2} \\\\ & =(10)^{2}-(6)^{2} \\\\ V_{R}^{2} & =100-36=64 \\\\ V_{R} & =8 \mathrm{~V} \end{aligned} $

In a series $L-C-R$ circuit where the current leads the voltage source, the reactance $(X)$ is predominantly capacitive, meaning the capacitive reactance ( $X_C$ ) is greater than the inductive reactane $\left(X_L\right)$.

Mathematically, it can be expressed as $X_C>X_L$ This imbalance the capacitive and inductive reactance values results in the leading current behaviour.

Given,

Coil resistance, $R=30 \Omega$

Inductive reactance, $X_L=20 \Omega$

Frequency $=50 \mathrm{~Hz}$

New source potential, $V=200 \mathrm{~V}$

New frequency $=100 \mathrm{~Hz}$

Inductive reactance is given by

$ \begin{aligned} & X_L=\omega L=2 \pi f L \\ & 20=\omega L=2 \pi \cdot 50 \times L \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

When $f=100 \mathrm{~Hz}=f^{\prime}$

$ \begin{aligned} X_L^{\prime} & =\omega^{\prime} L=2 \pi 100 L \\ X_L^{\prime} & =2 \pi(50 \times 2) \times L \\ X_L^{\prime} & =2 X_L=2 \times 20 \Omega \\ & =40 \Omega \end{aligned} $

Impedance of circuit.

$ \begin{aligned} &\begin{aligned} & Z=\sqrt{X_L^{\prime 2}+R^2} \\ & Z=\sqrt{(40)^2+(30)^2}=50 \Omega \end{aligned}\\ &\text { Using Ohm's Law }\\ &\begin{aligned} & i=\frac{V}{Z}=\frac{200}{50}=4 \mathrm{~A} \\ & i=4 \mathrm{~A} \end{aligned} \end{aligned} $

As, the frequency is very high, Inductance capacitance $X_L=2 \pi f L$, this will act as open circuit.

Capacitance reactance $X_C=\frac{1}{2 \pi f C}$, this will act as short circuit.

Hence, the resolved circuit is

Equivalent resistance is

$ 1 \Omega+\left(\frac{4 \times 4}{4+4}\right) \Omega+2 \Omega=5 \Omega $

Current in circuit is

$ \begin{array}{ll} & V=I R, I=\frac{V}{R} \\ \Rightarrow & I=\frac{220}{5}=44 \mathrm{~A} \end{array} $

Given $E=200 \sin (50 \pi t)$ volt

$ \begin{aligned} E_0 & =200 \mathrm{~V} \\ X_L & =40 \Omega \\ R & =30 \Omega \end{aligned} $

$ \begin{aligned}Impedance,\,\, Z & =\sqrt{R^2+X_L^2} \\ & =\sqrt{30^2+40^2}=50 \Omega \\ I_{\mathrm{rms}} & =\frac{E_{\mathrm{rms}}}{Z}=\frac{E_0}{\sqrt{2} Z}=\frac{200}{\sqrt{2} \times 50} \\ & =2 \sqrt{2} \mathrm{~A} \end{aligned} $

Power factor, $\cos \phi=\frac{R}{Z}=\frac{30}{50}=0.6$

Power dissipated

$ \begin{aligned} P & =E_{\mathrm{rms}} \times I_{\mathrm{rms}} \times \cos \phi \\ & =\frac{200}{\sqrt{2}} \times 2 \sqrt{2} \times 0.6 \\ & =240 \mathrm{~W} \end{aligned} $

$ \begin{aligned} &\text { Given, } V=150 \sin (80 \pi t) V\\ &\begin{aligned} R & =25 \Omega \Rightarrow V=75 \Omega \\ \text { Power factor } & =\cos \phi \\ & =\frac{R}{Z}=\frac{25}{75}=\frac{1}{3} \\ V_{\text {rms }} & =\frac{V_0}{\sqrt{2}}=\frac{150}{\sqrt{2}} \\ I_{\text {rms }} & =\frac{V_{\text {rms }}}{Z} \\ & =\frac{150}{75 \times \sqrt{2}}=\sqrt{2} \mathrm{~A} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { ∴ Average power dissipated per cycle, }\\ &\begin{aligned} P_{\mathrm{avg}} & =V_{\mathrm{rms}} \times I_{\mathrm{rms}} \times \cos \phi \\ & =\frac{150}{\sqrt{2}} \times \sqrt{2} \times \frac{1}{3}=50 \mathrm{~W} \end{aligned} \end{aligned} $