02. A.C. Circuit (L-C-R, LC Circuit)
Alternating Current

155160 A sinusoidal voltage with a frequency of $50 \mathrm{~Hz}$ is applied to a series LCR circuit with a resistance of $5 \Omega$ inductance of $20 \mathrm{mH}$ and a capacitance of $500 \mu \mathrm{F}$. The magnitude of impedance of the circuit is closed to

1 $19.2 \Omega$
2 $14.4 \Omega$
3 $9.6 \Omega$
4 Ans: d
Exp:D Given,
Resistance $(\mathrm{R})=5 \Omega$,
Inductance $(\mathrm{L})=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}$,
Capacitance $(\mathrm{C})=500 \mu \mathrm{F},=500 \times 10^{-6} \mathrm{~F}$,
Frequency $(\mathrm{f})=50 \mathrm{~Hz}$
Inductive reactance,
$\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}=2 \pi \times 50 \times 20 \times 10^{-3}=6.28 \Omega$
And, Capacitive reactance,
$\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \pi \times 50 \times 500 \times 10^{-6}}=\frac{1000}{50 \pi}=\frac{20}{\pi}=6.36 \Omega$
As we know that,
Impedance of LCR circuit,
$\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$
Then, $\quad Z=\sqrt{(5)^{2}+(6.28-6.36)^{2}}$
$=\sqrt{25+0.0064}=\sqrt{25.0064}$
$\mathrm{Z} \approx 5 \Omega$
Alternating Current

155161 A series $L C R$ circuit with $L=0.5 H$ and $R=$ $10 \Omega$ is connected to an AC supply with rms voltage and frequency equal to $200 \mathrm{~V}$ and $\frac{150}{\pi} \mathrm{Hz}$, respectively. The magnitude of the capacitance is varied so that current amplitude in the circuit becomes maximum. The rms voltage difference across the inductor is

1 $3000 \mathrm{~V}$
2 $2500 \mathrm{~V}$
3 $2000 \mathrm{~V}$
4 $2600 \mathrm{~V}$
Alternating Current

155162 An alternating voltage (in volt) given by $V=200 \sqrt{2} \sin (100 t)$ is connected to $1 \mu F$ capacitor through an $\mathrm{AC}$ ammeter. The reading of the ammeter will be

1 $10 \mathrm{~mA}$
2 $20 \mathrm{~mA}$
3 $40 \mathrm{~mA}$
4 $80 \mathrm{~mA}$
Alternating Current

155164 In a L-C circuit, angular frequency at resonance is $\omega$. What will be the new angular frequency when inductor's inductance is made two times and capacitor's capacitance is made four times?

1 $\frac{\omega}{2 \sqrt{2}}$
2 $\frac{\omega}{\sqrt{2}}$
3 $2 \omega$
4 $\frac{2 \omega}{\sqrt{2}}$
Alternating Current

155160 A sinusoidal voltage with a frequency of $50 \mathrm{~Hz}$ is applied to a series LCR circuit with a resistance of $5 \Omega$ inductance of $20 \mathrm{mH}$ and a capacitance of $500 \mu \mathrm{F}$. The magnitude of impedance of the circuit is closed to

1 $19.2 \Omega$
2 $14.4 \Omega$
3 $9.6 \Omega$
4 Ans: d
Exp:D Given,
Resistance $(\mathrm{R})=5 \Omega$,
Inductance $(\mathrm{L})=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}$,
Capacitance $(\mathrm{C})=500 \mu \mathrm{F},=500 \times 10^{-6} \mathrm{~F}$,
Frequency $(\mathrm{f})=50 \mathrm{~Hz}$
Inductive reactance,
$\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}=2 \pi \times 50 \times 20 \times 10^{-3}=6.28 \Omega$
And, Capacitive reactance,
$\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \pi \times 50 \times 500 \times 10^{-6}}=\frac{1000}{50 \pi}=\frac{20}{\pi}=6.36 \Omega$
As we know that,
Impedance of LCR circuit,
$\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$
Then, $\quad Z=\sqrt{(5)^{2}+(6.28-6.36)^{2}}$
$=\sqrt{25+0.0064}=\sqrt{25.0064}$
$\mathrm{Z} \approx 5 \Omega$
Alternating Current

155161 A series $L C R$ circuit with $L=0.5 H$ and $R=$ $10 \Omega$ is connected to an AC supply with rms voltage and frequency equal to $200 \mathrm{~V}$ and $\frac{150}{\pi} \mathrm{Hz}$, respectively. The magnitude of the capacitance is varied so that current amplitude in the circuit becomes maximum. The rms voltage difference across the inductor is

1 $3000 \mathrm{~V}$
2 $2500 \mathrm{~V}$
3 $2000 \mathrm{~V}$
4 $2600 \mathrm{~V}$
Alternating Current

155162 An alternating voltage (in volt) given by $V=200 \sqrt{2} \sin (100 t)$ is connected to $1 \mu F$ capacitor through an $\mathrm{AC}$ ammeter. The reading of the ammeter will be

1 $10 \mathrm{~mA}$
2 $20 \mathrm{~mA}$
3 $40 \mathrm{~mA}$
4 $80 \mathrm{~mA}$
Alternating Current

155164 In a L-C circuit, angular frequency at resonance is $\omega$. What will be the new angular frequency when inductor's inductance is made two times and capacitor's capacitance is made four times?

1 $\frac{\omega}{2 \sqrt{2}}$
2 $\frac{\omega}{\sqrt{2}}$
3 $2 \omega$
4 $\frac{2 \omega}{\sqrt{2}}$
Alternating Current

155160 A sinusoidal voltage with a frequency of $50 \mathrm{~Hz}$ is applied to a series LCR circuit with a resistance of $5 \Omega$ inductance of $20 \mathrm{mH}$ and a capacitance of $500 \mu \mathrm{F}$. The magnitude of impedance of the circuit is closed to

1 $19.2 \Omega$
2 $14.4 \Omega$
3 $9.6 \Omega$
4 Ans: d
Exp:D Given,
Resistance $(\mathrm{R})=5 \Omega$,
Inductance $(\mathrm{L})=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}$,
Capacitance $(\mathrm{C})=500 \mu \mathrm{F},=500 \times 10^{-6} \mathrm{~F}$,
Frequency $(\mathrm{f})=50 \mathrm{~Hz}$
Inductive reactance,
$\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}=2 \pi \times 50 \times 20 \times 10^{-3}=6.28 \Omega$
And, Capacitive reactance,
$\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \pi \times 50 \times 500 \times 10^{-6}}=\frac{1000}{50 \pi}=\frac{20}{\pi}=6.36 \Omega$
As we know that,
Impedance of LCR circuit,
$\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$
Then, $\quad Z=\sqrt{(5)^{2}+(6.28-6.36)^{2}}$
$=\sqrt{25+0.0064}=\sqrt{25.0064}$
$\mathrm{Z} \approx 5 \Omega$
Alternating Current

155161 A series $L C R$ circuit with $L=0.5 H$ and $R=$ $10 \Omega$ is connected to an AC supply with rms voltage and frequency equal to $200 \mathrm{~V}$ and $\frac{150}{\pi} \mathrm{Hz}$, respectively. The magnitude of the capacitance is varied so that current amplitude in the circuit becomes maximum. The rms voltage difference across the inductor is

1 $3000 \mathrm{~V}$
2 $2500 \mathrm{~V}$
3 $2000 \mathrm{~V}$
4 $2600 \mathrm{~V}$
Alternating Current

155162 An alternating voltage (in volt) given by $V=200 \sqrt{2} \sin (100 t)$ is connected to $1 \mu F$ capacitor through an $\mathrm{AC}$ ammeter. The reading of the ammeter will be

1 $10 \mathrm{~mA}$
2 $20 \mathrm{~mA}$
3 $40 \mathrm{~mA}$
4 $80 \mathrm{~mA}$
Alternating Current

155164 In a L-C circuit, angular frequency at resonance is $\omega$. What will be the new angular frequency when inductor's inductance is made two times and capacitor's capacitance is made four times?

1 $\frac{\omega}{2 \sqrt{2}}$
2 $\frac{\omega}{\sqrt{2}}$
3 $2 \omega$
4 $\frac{2 \omega}{\sqrt{2}}$
Alternating Current

155160 A sinusoidal voltage with a frequency of $50 \mathrm{~Hz}$ is applied to a series LCR circuit with a resistance of $5 \Omega$ inductance of $20 \mathrm{mH}$ and a capacitance of $500 \mu \mathrm{F}$. The magnitude of impedance of the circuit is closed to

1 $19.2 \Omega$
2 $14.4 \Omega$
3 $9.6 \Omega$
4 Ans: d
Exp:D Given,
Resistance $(\mathrm{R})=5 \Omega$,
Inductance $(\mathrm{L})=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}$,
Capacitance $(\mathrm{C})=500 \mu \mathrm{F},=500 \times 10^{-6} \mathrm{~F}$,
Frequency $(\mathrm{f})=50 \mathrm{~Hz}$
Inductive reactance,
$\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}=2 \pi \times 50 \times 20 \times 10^{-3}=6.28 \Omega$
And, Capacitive reactance,
$\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}=\frac{1}{2 \pi \times 50 \times 500 \times 10^{-6}}=\frac{1000}{50 \pi}=\frac{20}{\pi}=6.36 \Omega$
As we know that,
Impedance of LCR circuit,
$\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$
Then, $\quad Z=\sqrt{(5)^{2}+(6.28-6.36)^{2}}$
$=\sqrt{25+0.0064}=\sqrt{25.0064}$
$\mathrm{Z} \approx 5 \Omega$
Alternating Current

155161 A series $L C R$ circuit with $L=0.5 H$ and $R=$ $10 \Omega$ is connected to an AC supply with rms voltage and frequency equal to $200 \mathrm{~V}$ and $\frac{150}{\pi} \mathrm{Hz}$, respectively. The magnitude of the capacitance is varied so that current amplitude in the circuit becomes maximum. The rms voltage difference across the inductor is

1 $3000 \mathrm{~V}$
2 $2500 \mathrm{~V}$
3 $2000 \mathrm{~V}$
4 $2600 \mathrm{~V}$
Alternating Current

155162 An alternating voltage (in volt) given by $V=200 \sqrt{2} \sin (100 t)$ is connected to $1 \mu F$ capacitor through an $\mathrm{AC}$ ammeter. The reading of the ammeter will be

1 $10 \mathrm{~mA}$
2 $20 \mathrm{~mA}$
3 $40 \mathrm{~mA}$
4 $80 \mathrm{~mA}$
Alternating Current

155164 In a L-C circuit, angular frequency at resonance is $\omega$. What will be the new angular frequency when inductor's inductance is made two times and capacitor's capacitance is made four times?

1 $\frac{\omega}{2 \sqrt{2}}$
2 $\frac{\omega}{\sqrt{2}}$
3 $2 \omega$
4 $\frac{2 \omega}{\sqrt{2}}$