155221
The instantaneous values of current and voltage in an $A C$ circuit are $i=100 \sin 314 t$ amp and $e=200 \sin \left(314 t+\frac{\pi}{3}\right) V$ respectively. If the resistance is $1 \Omega$, then the reactance of the circuit will be-
1 $\sqrt{3} \Omega$
2 $100 \sqrt{3} \Omega$
3 $-200 \sqrt{3} \Omega$
4 $-200 / \sqrt{3} \Omega$
Explanation:
A Given that, Resistance $(\mathrm{R})=1 \Omega$ Current (i) $=100 \sin 314 \mathrm{t}$ amp. Voltage $(\mathrm{e})=200 \sin (314 \mathrm{t}+\pi / 3) \mathrm{v}$ $e_{0}=i_{0} Z$ $200=100 Z$ $Z=2 \Omega$ The reactance of the circuit is, $\mathrm{Z}^{2}=\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}$ $(2)^{2}=(1)^{2}+\mathrm{X}_{\mathrm{L}}^{2}$ $4=1+\mathrm{X}_{\mathrm{L}}^{2}$ $\mathrm{X}_{\mathrm{L}}^{2}=4-1$ $\mathrm{X}_{\mathrm{L}}=\sqrt{3} \Omega$
BCECE-2012
Alternating Current
155222
In a diode AM detector, the output circuit consists of $R=1 \mathrm{k} \Omega$ and $C=10 \mathrm{pF}$. A carrier signal of $100 \mathrm{k} \mathrm{Hz}$ is to be detected. Is it good?
1 Yes
2 No
3 Information is not sufficient
4 None of the above
Explanation:
B For better demodulation $\frac{1}{\mathrm{f}_{\mathrm{c}}} \lt \mathrm{RC}$ $\mathrm{R}=1 \mathrm{~K} \Omega, \mathrm{C}=10 \mathrm{pF}, \mathrm{f}_{\mathrm{c}}=100 \mathrm{kHz}=100 \times 10^{3} \mathrm{~Hz}$ $\frac{1}{\mathrm{f}_{\mathrm{c}}}=\frac{1}{100 \times 10^{3}}=10^{-5} \text { second }$ $\mathrm{RC}=1 \times 10^{3} \times 10 \times 10^{-12}=10^{-8} \text { second }$ $\frac{1}{\mathrm{f}_{\mathrm{c}}} \text { is more than } \mathrm{RC} \text { or } \frac{1}{\mathrm{f}_{\mathrm{c}}} \text { is not less than RC So it is }$ $\text { not good. }$
BCECE-2012
Alternating Current
155223
A series resonant circuit contains $L=\frac{5}{\pi} \mathrm{mH}$, $C=\frac{200}{\pi} \mu F$ and $R=100 \Omega$. If a source of emf e $=200 \sin 1000 \pi \mathrm{t}$ is applied, then the rms current is-
1 $2 \mathrm{~A}$
2 $200 \sqrt{2} \mathrm{~A}$
3 $100 \sqrt{2} \mathrm{~A}$
4 $1.41 \mathrm{~A}$
Explanation:
D Given, Capacitance $(C)=\frac{200}{\pi} \mu \mathrm{F}=\frac{200}{\pi} \times 10^{-6}$ Inductance $(\mathrm{L})=\frac{5}{\pi} \mathrm{mH}=\frac{5}{\pi} \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=100 \Omega$ Source of emf is given by $\mathrm{e}=200 \sin 1000 \pi \mathrm{t}$ On comparing with emf equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}$ is $\omega=1000 \pi$ $\mathrm{e}_{0}=200 \mathrm{v}$ At resonance, $\mathrm{Z}=\mathrm{R}$ So, current $\left(\mathrm{I}_{0}\right)=\frac{\mathrm{e}_{0}}{\mathrm{R}}=\frac{200}{100}=2 \mathrm{~A}$ Rms value of current $I_{\mathrm{rms}} =\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}$ $=\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2 \sqrt{2}}{2}$ $\mathrm{I}_{\mathrm{rms}} =\sqrt{2} \text { or } 1.41 \mathrm{Amp}$
BCECE-2011
Alternating Current
155224
A $100 \mathrm{~V}$, AC source of frequency $500 \mathrm{~Hz}$ is connected to an $L C R$ circuit with $L=8.1 \mathrm{mH}$, $C=12.5 \mu \mathrm{F}, \mathrm{R}=10 \Omega$ all connected in series as shown in figure. What is the quality factor of circuit?
155221
The instantaneous values of current and voltage in an $A C$ circuit are $i=100 \sin 314 t$ amp and $e=200 \sin \left(314 t+\frac{\pi}{3}\right) V$ respectively. If the resistance is $1 \Omega$, then the reactance of the circuit will be-
1 $\sqrt{3} \Omega$
2 $100 \sqrt{3} \Omega$
3 $-200 \sqrt{3} \Omega$
4 $-200 / \sqrt{3} \Omega$
Explanation:
A Given that, Resistance $(\mathrm{R})=1 \Omega$ Current (i) $=100 \sin 314 \mathrm{t}$ amp. Voltage $(\mathrm{e})=200 \sin (314 \mathrm{t}+\pi / 3) \mathrm{v}$ $e_{0}=i_{0} Z$ $200=100 Z$ $Z=2 \Omega$ The reactance of the circuit is, $\mathrm{Z}^{2}=\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}$ $(2)^{2}=(1)^{2}+\mathrm{X}_{\mathrm{L}}^{2}$ $4=1+\mathrm{X}_{\mathrm{L}}^{2}$ $\mathrm{X}_{\mathrm{L}}^{2}=4-1$ $\mathrm{X}_{\mathrm{L}}=\sqrt{3} \Omega$
BCECE-2012
Alternating Current
155222
In a diode AM detector, the output circuit consists of $R=1 \mathrm{k} \Omega$ and $C=10 \mathrm{pF}$. A carrier signal of $100 \mathrm{k} \mathrm{Hz}$ is to be detected. Is it good?
1 Yes
2 No
3 Information is not sufficient
4 None of the above
Explanation:
B For better demodulation $\frac{1}{\mathrm{f}_{\mathrm{c}}} \lt \mathrm{RC}$ $\mathrm{R}=1 \mathrm{~K} \Omega, \mathrm{C}=10 \mathrm{pF}, \mathrm{f}_{\mathrm{c}}=100 \mathrm{kHz}=100 \times 10^{3} \mathrm{~Hz}$ $\frac{1}{\mathrm{f}_{\mathrm{c}}}=\frac{1}{100 \times 10^{3}}=10^{-5} \text { second }$ $\mathrm{RC}=1 \times 10^{3} \times 10 \times 10^{-12}=10^{-8} \text { second }$ $\frac{1}{\mathrm{f}_{\mathrm{c}}} \text { is more than } \mathrm{RC} \text { or } \frac{1}{\mathrm{f}_{\mathrm{c}}} \text { is not less than RC So it is }$ $\text { not good. }$
BCECE-2012
Alternating Current
155223
A series resonant circuit contains $L=\frac{5}{\pi} \mathrm{mH}$, $C=\frac{200}{\pi} \mu F$ and $R=100 \Omega$. If a source of emf e $=200 \sin 1000 \pi \mathrm{t}$ is applied, then the rms current is-
1 $2 \mathrm{~A}$
2 $200 \sqrt{2} \mathrm{~A}$
3 $100 \sqrt{2} \mathrm{~A}$
4 $1.41 \mathrm{~A}$
Explanation:
D Given, Capacitance $(C)=\frac{200}{\pi} \mu \mathrm{F}=\frac{200}{\pi} \times 10^{-6}$ Inductance $(\mathrm{L})=\frac{5}{\pi} \mathrm{mH}=\frac{5}{\pi} \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=100 \Omega$ Source of emf is given by $\mathrm{e}=200 \sin 1000 \pi \mathrm{t}$ On comparing with emf equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}$ is $\omega=1000 \pi$ $\mathrm{e}_{0}=200 \mathrm{v}$ At resonance, $\mathrm{Z}=\mathrm{R}$ So, current $\left(\mathrm{I}_{0}\right)=\frac{\mathrm{e}_{0}}{\mathrm{R}}=\frac{200}{100}=2 \mathrm{~A}$ Rms value of current $I_{\mathrm{rms}} =\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}$ $=\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2 \sqrt{2}}{2}$ $\mathrm{I}_{\mathrm{rms}} =\sqrt{2} \text { or } 1.41 \mathrm{Amp}$
BCECE-2011
Alternating Current
155224
A $100 \mathrm{~V}$, AC source of frequency $500 \mathrm{~Hz}$ is connected to an $L C R$ circuit with $L=8.1 \mathrm{mH}$, $C=12.5 \mu \mathrm{F}, \mathrm{R}=10 \Omega$ all connected in series as shown in figure. What is the quality factor of circuit?
155221
The instantaneous values of current and voltage in an $A C$ circuit are $i=100 \sin 314 t$ amp and $e=200 \sin \left(314 t+\frac{\pi}{3}\right) V$ respectively. If the resistance is $1 \Omega$, then the reactance of the circuit will be-
1 $\sqrt{3} \Omega$
2 $100 \sqrt{3} \Omega$
3 $-200 \sqrt{3} \Omega$
4 $-200 / \sqrt{3} \Omega$
Explanation:
A Given that, Resistance $(\mathrm{R})=1 \Omega$ Current (i) $=100 \sin 314 \mathrm{t}$ amp. Voltage $(\mathrm{e})=200 \sin (314 \mathrm{t}+\pi / 3) \mathrm{v}$ $e_{0}=i_{0} Z$ $200=100 Z$ $Z=2 \Omega$ The reactance of the circuit is, $\mathrm{Z}^{2}=\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}$ $(2)^{2}=(1)^{2}+\mathrm{X}_{\mathrm{L}}^{2}$ $4=1+\mathrm{X}_{\mathrm{L}}^{2}$ $\mathrm{X}_{\mathrm{L}}^{2}=4-1$ $\mathrm{X}_{\mathrm{L}}=\sqrt{3} \Omega$
BCECE-2012
Alternating Current
155222
In a diode AM detector, the output circuit consists of $R=1 \mathrm{k} \Omega$ and $C=10 \mathrm{pF}$. A carrier signal of $100 \mathrm{k} \mathrm{Hz}$ is to be detected. Is it good?
1 Yes
2 No
3 Information is not sufficient
4 None of the above
Explanation:
B For better demodulation $\frac{1}{\mathrm{f}_{\mathrm{c}}} \lt \mathrm{RC}$ $\mathrm{R}=1 \mathrm{~K} \Omega, \mathrm{C}=10 \mathrm{pF}, \mathrm{f}_{\mathrm{c}}=100 \mathrm{kHz}=100 \times 10^{3} \mathrm{~Hz}$ $\frac{1}{\mathrm{f}_{\mathrm{c}}}=\frac{1}{100 \times 10^{3}}=10^{-5} \text { second }$ $\mathrm{RC}=1 \times 10^{3} \times 10 \times 10^{-12}=10^{-8} \text { second }$ $\frac{1}{\mathrm{f}_{\mathrm{c}}} \text { is more than } \mathrm{RC} \text { or } \frac{1}{\mathrm{f}_{\mathrm{c}}} \text { is not less than RC So it is }$ $\text { not good. }$
BCECE-2012
Alternating Current
155223
A series resonant circuit contains $L=\frac{5}{\pi} \mathrm{mH}$, $C=\frac{200}{\pi} \mu F$ and $R=100 \Omega$. If a source of emf e $=200 \sin 1000 \pi \mathrm{t}$ is applied, then the rms current is-
1 $2 \mathrm{~A}$
2 $200 \sqrt{2} \mathrm{~A}$
3 $100 \sqrt{2} \mathrm{~A}$
4 $1.41 \mathrm{~A}$
Explanation:
D Given, Capacitance $(C)=\frac{200}{\pi} \mu \mathrm{F}=\frac{200}{\pi} \times 10^{-6}$ Inductance $(\mathrm{L})=\frac{5}{\pi} \mathrm{mH}=\frac{5}{\pi} \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=100 \Omega$ Source of emf is given by $\mathrm{e}=200 \sin 1000 \pi \mathrm{t}$ On comparing with emf equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}$ is $\omega=1000 \pi$ $\mathrm{e}_{0}=200 \mathrm{v}$ At resonance, $\mathrm{Z}=\mathrm{R}$ So, current $\left(\mathrm{I}_{0}\right)=\frac{\mathrm{e}_{0}}{\mathrm{R}}=\frac{200}{100}=2 \mathrm{~A}$ Rms value of current $I_{\mathrm{rms}} =\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}$ $=\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2 \sqrt{2}}{2}$ $\mathrm{I}_{\mathrm{rms}} =\sqrt{2} \text { or } 1.41 \mathrm{Amp}$
BCECE-2011
Alternating Current
155224
A $100 \mathrm{~V}$, AC source of frequency $500 \mathrm{~Hz}$ is connected to an $L C R$ circuit with $L=8.1 \mathrm{mH}$, $C=12.5 \mu \mathrm{F}, \mathrm{R}=10 \Omega$ all connected in series as shown in figure. What is the quality factor of circuit?
155221
The instantaneous values of current and voltage in an $A C$ circuit are $i=100 \sin 314 t$ amp and $e=200 \sin \left(314 t+\frac{\pi}{3}\right) V$ respectively. If the resistance is $1 \Omega$, then the reactance of the circuit will be-
1 $\sqrt{3} \Omega$
2 $100 \sqrt{3} \Omega$
3 $-200 \sqrt{3} \Omega$
4 $-200 / \sqrt{3} \Omega$
Explanation:
A Given that, Resistance $(\mathrm{R})=1 \Omega$ Current (i) $=100 \sin 314 \mathrm{t}$ amp. Voltage $(\mathrm{e})=200 \sin (314 \mathrm{t}+\pi / 3) \mathrm{v}$ $e_{0}=i_{0} Z$ $200=100 Z$ $Z=2 \Omega$ The reactance of the circuit is, $\mathrm{Z}^{2}=\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}$ $(2)^{2}=(1)^{2}+\mathrm{X}_{\mathrm{L}}^{2}$ $4=1+\mathrm{X}_{\mathrm{L}}^{2}$ $\mathrm{X}_{\mathrm{L}}^{2}=4-1$ $\mathrm{X}_{\mathrm{L}}=\sqrt{3} \Omega$
BCECE-2012
Alternating Current
155222
In a diode AM detector, the output circuit consists of $R=1 \mathrm{k} \Omega$ and $C=10 \mathrm{pF}$. A carrier signal of $100 \mathrm{k} \mathrm{Hz}$ is to be detected. Is it good?
1 Yes
2 No
3 Information is not sufficient
4 None of the above
Explanation:
B For better demodulation $\frac{1}{\mathrm{f}_{\mathrm{c}}} \lt \mathrm{RC}$ $\mathrm{R}=1 \mathrm{~K} \Omega, \mathrm{C}=10 \mathrm{pF}, \mathrm{f}_{\mathrm{c}}=100 \mathrm{kHz}=100 \times 10^{3} \mathrm{~Hz}$ $\frac{1}{\mathrm{f}_{\mathrm{c}}}=\frac{1}{100 \times 10^{3}}=10^{-5} \text { second }$ $\mathrm{RC}=1 \times 10^{3} \times 10 \times 10^{-12}=10^{-8} \text { second }$ $\frac{1}{\mathrm{f}_{\mathrm{c}}} \text { is more than } \mathrm{RC} \text { or } \frac{1}{\mathrm{f}_{\mathrm{c}}} \text { is not less than RC So it is }$ $\text { not good. }$
BCECE-2012
Alternating Current
155223
A series resonant circuit contains $L=\frac{5}{\pi} \mathrm{mH}$, $C=\frac{200}{\pi} \mu F$ and $R=100 \Omega$. If a source of emf e $=200 \sin 1000 \pi \mathrm{t}$ is applied, then the rms current is-
1 $2 \mathrm{~A}$
2 $200 \sqrt{2} \mathrm{~A}$
3 $100 \sqrt{2} \mathrm{~A}$
4 $1.41 \mathrm{~A}$
Explanation:
D Given, Capacitance $(C)=\frac{200}{\pi} \mu \mathrm{F}=\frac{200}{\pi} \times 10^{-6}$ Inductance $(\mathrm{L})=\frac{5}{\pi} \mathrm{mH}=\frac{5}{\pi} \times 10^{-3} \mathrm{H}$ Resistance $(\mathrm{R})=100 \Omega$ Source of emf is given by $\mathrm{e}=200 \sin 1000 \pi \mathrm{t}$ On comparing with emf equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}$ is $\omega=1000 \pi$ $\mathrm{e}_{0}=200 \mathrm{v}$ At resonance, $\mathrm{Z}=\mathrm{R}$ So, current $\left(\mathrm{I}_{0}\right)=\frac{\mathrm{e}_{0}}{\mathrm{R}}=\frac{200}{100}=2 \mathrm{~A}$ Rms value of current $I_{\mathrm{rms}} =\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}$ $=\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2 \sqrt{2}}{2}$ $\mathrm{I}_{\mathrm{rms}} =\sqrt{2} \text { or } 1.41 \mathrm{Amp}$
BCECE-2011
Alternating Current
155224
A $100 \mathrm{~V}$, AC source of frequency $500 \mathrm{~Hz}$ is connected to an $L C R$ circuit with $L=8.1 \mathrm{mH}$, $C=12.5 \mu \mathrm{F}, \mathrm{R}=10 \Omega$ all connected in series as shown in figure. What is the quality factor of circuit?
