155351
The average power dissipated in $\mathrm{AC}$ circuit is 2 $W$. If a current flowing through a circuit is $2 \mathrm{~A}$ impedance is $1 \Omega$, what is the power factor of the AC circuit?
155352
In $\mathrm{R}-\mathrm{L}-\mathrm{C}$ series circuit, the potential differences across each element is $20 \mathrm{~V}$. Now the value of the resistance alone is doubled, then P.D. across $R, L$ and $C$ respectively.
A Given, potential across each element is same, so circuit at resonance. $\mathrm{VL}=\mathrm{VC}$ $\because$ If the value of resistance is doubled then current in the circuit is half of the initial value $I^{\prime}=\frac{I}{2}$ $V_{R}=20 \mathrm{~V}$ $V_{C}=V_{L}=\frac{X_{L} I}{2}=\frac{20}{2}=10 \mathrm{~V}$
Karnataka CET-2013
Alternating Current
155353
In an $\mathrm{AC}$ circuit the instantaneous values of emf and current are $V=200 \sin 300 t$ volt and $i=2 \sin \left(300 t+\frac{\pi}{3}\right) \text { ampere }$ the average power consumed in watts is
155354
In an A.C. circuit the instantaneous values of emf and current are $E=200 \sin 314 t$ volt and $I=\frac{1}{2} \sin \left(314 t+\frac{\pi}{3}\right)$ ampere. The average power consumed in watts is
1 200
2 100
3 50
4 25
Explanation:
D Given that, $E=200 \sin (314 t) \text { volts }$ $I=\frac{1}{2} \sin \left(314 t+\frac{\pi}{3}\right) A$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t})$ and $\mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}$ $+\phi)$ $\text { Then, } \mathrm{I}_{0} =\frac{1}{2} \mathrm{~A}$ $\mathrm{E}_{0} =200 \mathrm{~V}$ $\phi =\frac{\pi}{3}=60^{\circ}$ The average power consumed $\mathrm{P}_{\mathrm{avg}}=\frac{\mathrm{I}_{0} \cdot \mathrm{E}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}}=\frac{200}{2} \times \frac{1}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}}=50 \times \frac{1}{2}$ $\mathrm{P}_{\mathrm{avg}}=25 \mathrm{~W}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155351
The average power dissipated in $\mathrm{AC}$ circuit is 2 $W$. If a current flowing through a circuit is $2 \mathrm{~A}$ impedance is $1 \Omega$, what is the power factor of the AC circuit?
155352
In $\mathrm{R}-\mathrm{L}-\mathrm{C}$ series circuit, the potential differences across each element is $20 \mathrm{~V}$. Now the value of the resistance alone is doubled, then P.D. across $R, L$ and $C$ respectively.
A Given, potential across each element is same, so circuit at resonance. $\mathrm{VL}=\mathrm{VC}$ $\because$ If the value of resistance is doubled then current in the circuit is half of the initial value $I^{\prime}=\frac{I}{2}$ $V_{R}=20 \mathrm{~V}$ $V_{C}=V_{L}=\frac{X_{L} I}{2}=\frac{20}{2}=10 \mathrm{~V}$
Karnataka CET-2013
Alternating Current
155353
In an $\mathrm{AC}$ circuit the instantaneous values of emf and current are $V=200 \sin 300 t$ volt and $i=2 \sin \left(300 t+\frac{\pi}{3}\right) \text { ampere }$ the average power consumed in watts is
155354
In an A.C. circuit the instantaneous values of emf and current are $E=200 \sin 314 t$ volt and $I=\frac{1}{2} \sin \left(314 t+\frac{\pi}{3}\right)$ ampere. The average power consumed in watts is
1 200
2 100
3 50
4 25
Explanation:
D Given that, $E=200 \sin (314 t) \text { volts }$ $I=\frac{1}{2} \sin \left(314 t+\frac{\pi}{3}\right) A$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t})$ and $\mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}$ $+\phi)$ $\text { Then, } \mathrm{I}_{0} =\frac{1}{2} \mathrm{~A}$ $\mathrm{E}_{0} =200 \mathrm{~V}$ $\phi =\frac{\pi}{3}=60^{\circ}$ The average power consumed $\mathrm{P}_{\mathrm{avg}}=\frac{\mathrm{I}_{0} \cdot \mathrm{E}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}}=\frac{200}{2} \times \frac{1}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}}=50 \times \frac{1}{2}$ $\mathrm{P}_{\mathrm{avg}}=25 \mathrm{~W}$
155351
The average power dissipated in $\mathrm{AC}$ circuit is 2 $W$. If a current flowing through a circuit is $2 \mathrm{~A}$ impedance is $1 \Omega$, what is the power factor of the AC circuit?
155352
In $\mathrm{R}-\mathrm{L}-\mathrm{C}$ series circuit, the potential differences across each element is $20 \mathrm{~V}$. Now the value of the resistance alone is doubled, then P.D. across $R, L$ and $C$ respectively.
A Given, potential across each element is same, so circuit at resonance. $\mathrm{VL}=\mathrm{VC}$ $\because$ If the value of resistance is doubled then current in the circuit is half of the initial value $I^{\prime}=\frac{I}{2}$ $V_{R}=20 \mathrm{~V}$ $V_{C}=V_{L}=\frac{X_{L} I}{2}=\frac{20}{2}=10 \mathrm{~V}$
Karnataka CET-2013
Alternating Current
155353
In an $\mathrm{AC}$ circuit the instantaneous values of emf and current are $V=200 \sin 300 t$ volt and $i=2 \sin \left(300 t+\frac{\pi}{3}\right) \text { ampere }$ the average power consumed in watts is
155354
In an A.C. circuit the instantaneous values of emf and current are $E=200 \sin 314 t$ volt and $I=\frac{1}{2} \sin \left(314 t+\frac{\pi}{3}\right)$ ampere. The average power consumed in watts is
1 200
2 100
3 50
4 25
Explanation:
D Given that, $E=200 \sin (314 t) \text { volts }$ $I=\frac{1}{2} \sin \left(314 t+\frac{\pi}{3}\right) A$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t})$ and $\mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}$ $+\phi)$ $\text { Then, } \mathrm{I}_{0} =\frac{1}{2} \mathrm{~A}$ $\mathrm{E}_{0} =200 \mathrm{~V}$ $\phi =\frac{\pi}{3}=60^{\circ}$ The average power consumed $\mathrm{P}_{\mathrm{avg}}=\frac{\mathrm{I}_{0} \cdot \mathrm{E}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}}=\frac{200}{2} \times \frac{1}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}}=50 \times \frac{1}{2}$ $\mathrm{P}_{\mathrm{avg}}=25 \mathrm{~W}$
155351
The average power dissipated in $\mathrm{AC}$ circuit is 2 $W$. If a current flowing through a circuit is $2 \mathrm{~A}$ impedance is $1 \Omega$, what is the power factor of the AC circuit?
155352
In $\mathrm{R}-\mathrm{L}-\mathrm{C}$ series circuit, the potential differences across each element is $20 \mathrm{~V}$. Now the value of the resistance alone is doubled, then P.D. across $R, L$ and $C$ respectively.
A Given, potential across each element is same, so circuit at resonance. $\mathrm{VL}=\mathrm{VC}$ $\because$ If the value of resistance is doubled then current in the circuit is half of the initial value $I^{\prime}=\frac{I}{2}$ $V_{R}=20 \mathrm{~V}$ $V_{C}=V_{L}=\frac{X_{L} I}{2}=\frac{20}{2}=10 \mathrm{~V}$
Karnataka CET-2013
Alternating Current
155353
In an $\mathrm{AC}$ circuit the instantaneous values of emf and current are $V=200 \sin 300 t$ volt and $i=2 \sin \left(300 t+\frac{\pi}{3}\right) \text { ampere }$ the average power consumed in watts is
155354
In an A.C. circuit the instantaneous values of emf and current are $E=200 \sin 314 t$ volt and $I=\frac{1}{2} \sin \left(314 t+\frac{\pi}{3}\right)$ ampere. The average power consumed in watts is
1 200
2 100
3 50
4 25
Explanation:
D Given that, $E=200 \sin (314 t) \text { volts }$ $I=\frac{1}{2} \sin \left(314 t+\frac{\pi}{3}\right) A$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t})$ and $\mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}$ $+\phi)$ $\text { Then, } \mathrm{I}_{0} =\frac{1}{2} \mathrm{~A}$ $\mathrm{E}_{0} =200 \mathrm{~V}$ $\phi =\frac{\pi}{3}=60^{\circ}$ The average power consumed $\mathrm{P}_{\mathrm{avg}}=\frac{\mathrm{I}_{0} \cdot \mathrm{E}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}}=\frac{200}{2} \times \frac{1}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}}=50 \times \frac{1}{2}$ $\mathrm{P}_{\mathrm{avg}}=25 \mathrm{~W}$