Answer

Resistance of the resistor, $R=100 \Omega$

Supply voltage, $V=220 \mathrm{~V}$

Frequency, $v=50 \mathrm{~Hz}$

(a) The rms value of current in the circuit is given as:

$$ \begin{aligned} I & =\frac{V}{R} \\ & =\frac{220}{100}=2.20 \mathrm{~A} \end{aligned} $$

(b) The net power consumed over a full cycle is given as:

$$ P=V I $$

$=220 \times 2.2=484 \mathrm{~W}$

Answer

(a) Peak voltage of the ac supply, $V_{0}=300 \mathrm{~V}$

Rms voltage is given as:

$$ \begin{aligned} V & =\frac{V_{0}}{\sqrt{2}} \\ & =\frac{300}{\sqrt{2}}=212.1 \mathrm{~V} \end{aligned} $$

(b) Therms value of current is given as:

$I=10 \mathrm{~A}$

Now, peak current is given as:

$$ \begin{aligned} I_{0} & =\sqrt{2} I \\ & =10 \sqrt{2}=14.1 \mathrm{~A} \end{aligned} $$

Answer

Inductance of inductor, $L=44 \mathrm{mH}=44 \times 10^{-3} \mathrm{H}$

Supply voltage, $V=220 \mathrm{~V}$

Frequency, $v=50 \mathrm{~Hz}$

Angular frequency, $\omega=2 \pi v$

Inductive reactance, $X_{\mathrm{L}}=\omega L=2 \pi \nu L=2 \pi \times 50 \times 44 \times 10^{-3} \Omega$

Rms value of current is given as:

$$ \begin{aligned} I & =\frac{v}{X_{\mathrm{L}}} \\ & =\frac{220}{2 \pi \times 50 \times 44 \times 10^{-3}}=15.92 \mathrm{~A} \end{aligned} $$

Hence, the rms value of current in the circuit is $15.92 \mathrm{~A}$.

Answer

Capacitance of capacitor, $C=60 \mu \mathrm{F}=60 \times 10^{-6} \mathrm{~F}$

Supply voltage, $V=110 \mathrm{~V}$

Frequency, $v=60 \mathrm{~Hz}$

Angular frequency, $\omega=2 \pi v$

Capacitive reactance $X_{\mathrm{c}}=\frac{1}{\omega C}$

$=\frac{1}{2 \pi v C}$

$=\frac{1}{2 \times 3.14 \times 60 \times 60 \times 10^{-6}} \Omega^{-1}$

Rms value of current is given as:

$$ \begin{aligned} I & =\frac{v}{X_{\mathrm{c}}} \\ & =110 \times 2 \times 3.14 \times 60 \times 10^{-6} \times 60=2.49 \mathrm{~A} \end{aligned} $$

Hence, the rms value of current is $2.49 \mathrm{~A}$.

Answer

In the inductive circuit,

Rms value of current, $I=15.92 \mathrm{~A}$

Rms value of voltage, $V=220 \mathrm{~V}$

Hence, the net power absorbed can be obtained by the relation,

$P=V I \cos \Phi$

Where,

$\Phi=$ Phase difference between $V$ and $I$

For a pure inductive circuit, the phase difference between alternating voltage and current is $90^{\circ}$ i.e., $\Phi=90^{\circ}$.

Hence, $P=0$ i.e., the net power is zero.

In the capacitive circuit,

Rms value of current, $I=2.49$ A

Rms value of voltage, $V=110 \mathrm{~V}$

Hence, the net power absorbed can ve obtained as:

$P=V I \operatorname{Cos} \Phi$

For a pure capacitive circuit, the phase difference between alternating voltage and current is $90^{\circ}$ i.e., $\Phi=90^{\circ}$.

Hence, $P=0$ i.e., the net power is zero.

Answer

Capacitance, $C=30 \mu \mathrm{F}=30 \times 10^{-6} \mathrm{~F}$

Inductance, $L=27 \mathrm{mH}=27 \times 10^{-3} \mathrm{H}$

Angular frequency is given as:

$$ \begin{aligned} \omega_{r} & =\frac{1}{\sqrt{L C}} \\ & =\frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}}=\frac{1}{9 \times 10^{-4}}=1.11 \times 10^{3} \mathrm{rad} / \mathrm{s} \end{aligned} $$

Hence, the angular frequency of free oscillations of the circuit is $1.11 \times 10^{3} \mathrm{rad} / \mathrm{s}$.

Answer

At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.

Resistance, $R=20 \Omega$

Inductance, $L=1.5 \mathrm{H}$

Capacitance, $C=35 \mu \mathrm{F}=30 \times 10^{-6} \mathrm{~F}$

AC supply voltage to the $L C R$ circuit, $V=200 \mathrm{~V}$

Impedance of the circuit is given by the relation,

$Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$

At resonance, $\omega L=\frac{1}{\omega C}$

$\therefore Z=R=20 \Omega$

Current in the circuit can be calculated as:

$$ \begin{aligned} I & =\frac{V}{Z} \\ & =\frac{200}{20}=10 \mathrm{~A} \end{aligned} $$

Hence, the average power transferred to the circuit in one complete cycle $=V I$

$=200 \times 10=2000 \mathrm{~W}$.

Answer

Inductance of the inductor, $L=5.0 \mathrm{H}$ Capacitance of the capacitor, $C=80 \mu \mathrm{H}=80 \times 10^{-6} \mathrm{~F}$ Resistance of the resistor, $R=40 \Omega$ Potential of the variable voltage source, $V=230 \mathrm{~V}$ (a) Resonance angular frequency is given as: $$ \begin{aligned} \omega_R & =\frac{1}{\sqrt{L C}} \ & =\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}=\frac{10^3}{20}=50 \mathrm{rad} / \mathrm{s} \end{aligned} $$

Hence, the circuit will come in resonance for a source frequency of $50 \mathrm{rad} / \mathrm{s}$.

(b) Impedance of the circuit is given by the relation, $$ Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} $$

At resonance, $$ \begin{aligned} & \omega L=\frac{1}{\omega C} \ & \therefore Z=R=40 \Omega \end{aligned} $$

Amplitude of the current at the resonating frequency is given as: $I_0=\frac{V_0}{Z}$ Where, $$ \begin{aligned} V_0 & =\text { Peak voltage } \ & =\sqrt{2} \mathrm{~V} \ \therefore I_0 & =\frac{\sqrt{2} \mathrm{~V}}{Z} \ & =\frac{\sqrt{2} \times 230}{40}=8.13 \mathrm{~A} \end{aligned} $$

Hence, at resonance, the impedance of the circuit is $40 \Omega$ and the amplitude of the current is $8.13 \mathrm{~A}$.

(c) Rms potential drop across the inductor, $$ \left(V_L\right)_{\text {rms }}=I \times \omega_R L $$

Where, $I=$ rms current $$ \begin{aligned} & =\frac{I_0}{\sqrt{2}}=\frac{\sqrt{2} V}{\sqrt{2} Z}=\frac{230}{40} \mathrm{~A} \ & \therefore\left(V_L\right)_{\text {rms }}=\frac{230}{40} \times 50 \times 5=1437.5 \mathrm{~V} \end{aligned} $$

Potential drop across the capacitor, $$ \begin{aligned} \left(V_c\right)_{\mathrm{ms}} & =I \times \frac{1}{\omega_R C} \ & =\frac{230}{40} \times \frac{1}{50 \times 80 \times 10^{-6}}=1437.5 \mathrm{~V} \end{aligned} $$

Potential drop across the resistor, $$ \begin{aligned} & \left(V_R\right)_{\mathrm{rms}}=I R \ & =\frac{230}{40} \times 40=230 \mathrm{~V} \end{aligned} $$

Potential drop across the LC combination, $$ V_{L C}=I\left(\omega_R L-\frac{1}{\omega_R C}\right) $$

At resonance, $\omega_R L=\frac{1}{\omega_R C}$ $$ \therefore V_{L C}=0 $$

Hence, it is proved that the potential drop across the $L C$ combination is zero at resonating frequency.