155345
In an A.C. circuit, the current flowing in inductance is $I=5 \sin (100 t-\pi / 2)$ amperes and the potential difference is $V=200 \sin (100 t)$ volts. The power consumption is equal to
155346
In a $\mathrm{AC}$ circuit the voltage and current are described by $V=200 \sin \left(319 t-\frac{\pi}{6}\right)$ volts and $I=50 \sin \left(314 t+\frac{\pi}{6}\right) \mathrm{mA}$ respectively. The average power dissipated in the circuit is:
1 2.5 watts
2 5.0 watts
3 10.0 watts
4 50.0 watts
Explanation:
A Given, $\mathrm{V}=200 \sin \left(319 \mathrm{t}-\frac{\pi}{6}\right) \mathrm{V}$ $\mathrm{I}=50 \sin \left(314 \mathrm{t}+\frac{\pi}{6}\right) \mathrm{mA}$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t}+\phi)$ and $\mathrm{I}=\mathrm{I}_{0} \sin$ $(\omega \mathrm{t}+\phi)$ Then, $\mathrm{V}_{0}=200 \mathrm{~V}, \mathrm{I}_{0}=50 \mathrm{~mA}=50 \times 10^{-3} \mathrm{~A}$ $\phi=\phi_{1}-\phi_{2}=\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)=\frac{\pi}{3}$ The average power dissipated in the circuit - $\mathrm{P}_{\mathrm{avg}} =\frac{\mathrm{V}_{0} \mathrm{I}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}} =\frac{200 \times 50 \times 10^{-3}}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}} =2.5 \mathrm{~W}$
AIIMS-2011
Alternating Current
155347
An electric power station transmits $100 \mathrm{MW}$ power through long and thin cable. If the transmission is at (i) $20000 \mathrm{~V}$, (ii) $200 \mathrm{~V}$, in which case would be less power loss?
1 In (i) only
2 In (ii) only
3 In each case, power loss is zero
4 Data is insufficient
Explanation:
A Given, power of electric station $\mathrm{P}=100 \mathrm{MW}=100 \times 10^{6} \mathrm{~W}=10^{8} \mathrm{~W}$ (i) Given, $\mathrm{V}_{1}=20,000 \mathrm{~V}$ $I_{1}=\frac{P}{V_{1}}=\frac{10^{8}}{20,000}=\frac{10^{8}}{2 \times 10^{4}}=\frac{10000}{2}$ $I_{1}=5000 \mathrm{~A}$ Then rate of heat dissipation, $\mathrm{P}_{1}=\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{P}_{1}=(5000)^{2} \mathrm{R}$ $\mathrm{P}_{1}=25 \times 10^{6} \mathrm{R} \mathrm{W}$ (ii) Given, $\mathrm{V}_{2}=200 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{\mathrm{P}}{\mathrm{V}_{2}}=\frac{10^{8}}{200}=5 \times 10^{5} \mathrm{~A}$ $\mathrm{P}_{2}=\mathrm{I}_{2}^{2} \mathrm{R}=\left(5 \times 10^{5}\right)^{2} \mathrm{R}$ $\mathrm{P}_{2}=25 \times 10^{10} \mathrm{RW}$ Comparing equation (i) and equation (ii), we get - There will be lesser power wastage when power is transmitted at 20,000 V
BCECE-2004
Alternating Current
155349
In an $\mathrm{AC}$ circuit, the instantaneous values of emf and current are $V=200 \sin (314 t) V$ and $I=\sin \left(314 t+\frac{\pi}{3}\right) A . \quad$ The average power consumed (in W) is
155350
The average power dissipated in a pure inductor is :
1 $\frac{\mathrm{VI}^{2}}{4}$
2 $\frac{1}{2} \mathrm{VI}$
3 Zero
4 $\mathrm{VI}^{2}$
Explanation:
C The average power dissipated is $\mathrm{P}_{\mathrm{avg}}=\mathrm{VI} \cos \phi$ Where, $\cos \phi$ is the power factor. For pure inductor, $\phi=\frac{\pi}{2}$ $\therefore \operatorname{Cos} \phi$ or $\mathrm{P}_{\text {avg }}=0$ Hence, there is no power dissipation in pure inductor.
155345
In an A.C. circuit, the current flowing in inductance is $I=5 \sin (100 t-\pi / 2)$ amperes and the potential difference is $V=200 \sin (100 t)$ volts. The power consumption is equal to
155346
In a $\mathrm{AC}$ circuit the voltage and current are described by $V=200 \sin \left(319 t-\frac{\pi}{6}\right)$ volts and $I=50 \sin \left(314 t+\frac{\pi}{6}\right) \mathrm{mA}$ respectively. The average power dissipated in the circuit is:
1 2.5 watts
2 5.0 watts
3 10.0 watts
4 50.0 watts
Explanation:
A Given, $\mathrm{V}=200 \sin \left(319 \mathrm{t}-\frac{\pi}{6}\right) \mathrm{V}$ $\mathrm{I}=50 \sin \left(314 \mathrm{t}+\frac{\pi}{6}\right) \mathrm{mA}$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t}+\phi)$ and $\mathrm{I}=\mathrm{I}_{0} \sin$ $(\omega \mathrm{t}+\phi)$ Then, $\mathrm{V}_{0}=200 \mathrm{~V}, \mathrm{I}_{0}=50 \mathrm{~mA}=50 \times 10^{-3} \mathrm{~A}$ $\phi=\phi_{1}-\phi_{2}=\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)=\frac{\pi}{3}$ The average power dissipated in the circuit - $\mathrm{P}_{\mathrm{avg}} =\frac{\mathrm{V}_{0} \mathrm{I}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}} =\frac{200 \times 50 \times 10^{-3}}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}} =2.5 \mathrm{~W}$
AIIMS-2011
Alternating Current
155347
An electric power station transmits $100 \mathrm{MW}$ power through long and thin cable. If the transmission is at (i) $20000 \mathrm{~V}$, (ii) $200 \mathrm{~V}$, in which case would be less power loss?
1 In (i) only
2 In (ii) only
3 In each case, power loss is zero
4 Data is insufficient
Explanation:
A Given, power of electric station $\mathrm{P}=100 \mathrm{MW}=100 \times 10^{6} \mathrm{~W}=10^{8} \mathrm{~W}$ (i) Given, $\mathrm{V}_{1}=20,000 \mathrm{~V}$ $I_{1}=\frac{P}{V_{1}}=\frac{10^{8}}{20,000}=\frac{10^{8}}{2 \times 10^{4}}=\frac{10000}{2}$ $I_{1}=5000 \mathrm{~A}$ Then rate of heat dissipation, $\mathrm{P}_{1}=\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{P}_{1}=(5000)^{2} \mathrm{R}$ $\mathrm{P}_{1}=25 \times 10^{6} \mathrm{R} \mathrm{W}$ (ii) Given, $\mathrm{V}_{2}=200 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{\mathrm{P}}{\mathrm{V}_{2}}=\frac{10^{8}}{200}=5 \times 10^{5} \mathrm{~A}$ $\mathrm{P}_{2}=\mathrm{I}_{2}^{2} \mathrm{R}=\left(5 \times 10^{5}\right)^{2} \mathrm{R}$ $\mathrm{P}_{2}=25 \times 10^{10} \mathrm{RW}$ Comparing equation (i) and equation (ii), we get - There will be lesser power wastage when power is transmitted at 20,000 V
BCECE-2004
Alternating Current
155349
In an $\mathrm{AC}$ circuit, the instantaneous values of emf and current are $V=200 \sin (314 t) V$ and $I=\sin \left(314 t+\frac{\pi}{3}\right) A . \quad$ The average power consumed (in W) is
155350
The average power dissipated in a pure inductor is :
1 $\frac{\mathrm{VI}^{2}}{4}$
2 $\frac{1}{2} \mathrm{VI}$
3 Zero
4 $\mathrm{VI}^{2}$
Explanation:
C The average power dissipated is $\mathrm{P}_{\mathrm{avg}}=\mathrm{VI} \cos \phi$ Where, $\cos \phi$ is the power factor. For pure inductor, $\phi=\frac{\pi}{2}$ $\therefore \operatorname{Cos} \phi$ or $\mathrm{P}_{\text {avg }}=0$ Hence, there is no power dissipation in pure inductor.
155345
In an A.C. circuit, the current flowing in inductance is $I=5 \sin (100 t-\pi / 2)$ amperes and the potential difference is $V=200 \sin (100 t)$ volts. The power consumption is equal to
155346
In a $\mathrm{AC}$ circuit the voltage and current are described by $V=200 \sin \left(319 t-\frac{\pi}{6}\right)$ volts and $I=50 \sin \left(314 t+\frac{\pi}{6}\right) \mathrm{mA}$ respectively. The average power dissipated in the circuit is:
1 2.5 watts
2 5.0 watts
3 10.0 watts
4 50.0 watts
Explanation:
A Given, $\mathrm{V}=200 \sin \left(319 \mathrm{t}-\frac{\pi}{6}\right) \mathrm{V}$ $\mathrm{I}=50 \sin \left(314 \mathrm{t}+\frac{\pi}{6}\right) \mathrm{mA}$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t}+\phi)$ and $\mathrm{I}=\mathrm{I}_{0} \sin$ $(\omega \mathrm{t}+\phi)$ Then, $\mathrm{V}_{0}=200 \mathrm{~V}, \mathrm{I}_{0}=50 \mathrm{~mA}=50 \times 10^{-3} \mathrm{~A}$ $\phi=\phi_{1}-\phi_{2}=\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)=\frac{\pi}{3}$ The average power dissipated in the circuit - $\mathrm{P}_{\mathrm{avg}} =\frac{\mathrm{V}_{0} \mathrm{I}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}} =\frac{200 \times 50 \times 10^{-3}}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}} =2.5 \mathrm{~W}$
AIIMS-2011
Alternating Current
155347
An electric power station transmits $100 \mathrm{MW}$ power through long and thin cable. If the transmission is at (i) $20000 \mathrm{~V}$, (ii) $200 \mathrm{~V}$, in which case would be less power loss?
1 In (i) only
2 In (ii) only
3 In each case, power loss is zero
4 Data is insufficient
Explanation:
A Given, power of electric station $\mathrm{P}=100 \mathrm{MW}=100 \times 10^{6} \mathrm{~W}=10^{8} \mathrm{~W}$ (i) Given, $\mathrm{V}_{1}=20,000 \mathrm{~V}$ $I_{1}=\frac{P}{V_{1}}=\frac{10^{8}}{20,000}=\frac{10^{8}}{2 \times 10^{4}}=\frac{10000}{2}$ $I_{1}=5000 \mathrm{~A}$ Then rate of heat dissipation, $\mathrm{P}_{1}=\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{P}_{1}=(5000)^{2} \mathrm{R}$ $\mathrm{P}_{1}=25 \times 10^{6} \mathrm{R} \mathrm{W}$ (ii) Given, $\mathrm{V}_{2}=200 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{\mathrm{P}}{\mathrm{V}_{2}}=\frac{10^{8}}{200}=5 \times 10^{5} \mathrm{~A}$ $\mathrm{P}_{2}=\mathrm{I}_{2}^{2} \mathrm{R}=\left(5 \times 10^{5}\right)^{2} \mathrm{R}$ $\mathrm{P}_{2}=25 \times 10^{10} \mathrm{RW}$ Comparing equation (i) and equation (ii), we get - There will be lesser power wastage when power is transmitted at 20,000 V
BCECE-2004
Alternating Current
155349
In an $\mathrm{AC}$ circuit, the instantaneous values of emf and current are $V=200 \sin (314 t) V$ and $I=\sin \left(314 t+\frac{\pi}{3}\right) A . \quad$ The average power consumed (in W) is
155350
The average power dissipated in a pure inductor is :
1 $\frac{\mathrm{VI}^{2}}{4}$
2 $\frac{1}{2} \mathrm{VI}$
3 Zero
4 $\mathrm{VI}^{2}$
Explanation:
C The average power dissipated is $\mathrm{P}_{\mathrm{avg}}=\mathrm{VI} \cos \phi$ Where, $\cos \phi$ is the power factor. For pure inductor, $\phi=\frac{\pi}{2}$ $\therefore \operatorname{Cos} \phi$ or $\mathrm{P}_{\text {avg }}=0$ Hence, there is no power dissipation in pure inductor.
155345
In an A.C. circuit, the current flowing in inductance is $I=5 \sin (100 t-\pi / 2)$ amperes and the potential difference is $V=200 \sin (100 t)$ volts. The power consumption is equal to
155346
In a $\mathrm{AC}$ circuit the voltage and current are described by $V=200 \sin \left(319 t-\frac{\pi}{6}\right)$ volts and $I=50 \sin \left(314 t+\frac{\pi}{6}\right) \mathrm{mA}$ respectively. The average power dissipated in the circuit is:
1 2.5 watts
2 5.0 watts
3 10.0 watts
4 50.0 watts
Explanation:
A Given, $\mathrm{V}=200 \sin \left(319 \mathrm{t}-\frac{\pi}{6}\right) \mathrm{V}$ $\mathrm{I}=50 \sin \left(314 \mathrm{t}+\frac{\pi}{6}\right) \mathrm{mA}$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t}+\phi)$ and $\mathrm{I}=\mathrm{I}_{0} \sin$ $(\omega \mathrm{t}+\phi)$ Then, $\mathrm{V}_{0}=200 \mathrm{~V}, \mathrm{I}_{0}=50 \mathrm{~mA}=50 \times 10^{-3} \mathrm{~A}$ $\phi=\phi_{1}-\phi_{2}=\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)=\frac{\pi}{3}$ The average power dissipated in the circuit - $\mathrm{P}_{\mathrm{avg}} =\frac{\mathrm{V}_{0} \mathrm{I}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}} =\frac{200 \times 50 \times 10^{-3}}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}} =2.5 \mathrm{~W}$
AIIMS-2011
Alternating Current
155347
An electric power station transmits $100 \mathrm{MW}$ power through long and thin cable. If the transmission is at (i) $20000 \mathrm{~V}$, (ii) $200 \mathrm{~V}$, in which case would be less power loss?
1 In (i) only
2 In (ii) only
3 In each case, power loss is zero
4 Data is insufficient
Explanation:
A Given, power of electric station $\mathrm{P}=100 \mathrm{MW}=100 \times 10^{6} \mathrm{~W}=10^{8} \mathrm{~W}$ (i) Given, $\mathrm{V}_{1}=20,000 \mathrm{~V}$ $I_{1}=\frac{P}{V_{1}}=\frac{10^{8}}{20,000}=\frac{10^{8}}{2 \times 10^{4}}=\frac{10000}{2}$ $I_{1}=5000 \mathrm{~A}$ Then rate of heat dissipation, $\mathrm{P}_{1}=\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{P}_{1}=(5000)^{2} \mathrm{R}$ $\mathrm{P}_{1}=25 \times 10^{6} \mathrm{R} \mathrm{W}$ (ii) Given, $\mathrm{V}_{2}=200 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{\mathrm{P}}{\mathrm{V}_{2}}=\frac{10^{8}}{200}=5 \times 10^{5} \mathrm{~A}$ $\mathrm{P}_{2}=\mathrm{I}_{2}^{2} \mathrm{R}=\left(5 \times 10^{5}\right)^{2} \mathrm{R}$ $\mathrm{P}_{2}=25 \times 10^{10} \mathrm{RW}$ Comparing equation (i) and equation (ii), we get - There will be lesser power wastage when power is transmitted at 20,000 V
BCECE-2004
Alternating Current
155349
In an $\mathrm{AC}$ circuit, the instantaneous values of emf and current are $V=200 \sin (314 t) V$ and $I=\sin \left(314 t+\frac{\pi}{3}\right) A . \quad$ The average power consumed (in W) is
155350
The average power dissipated in a pure inductor is :
1 $\frac{\mathrm{VI}^{2}}{4}$
2 $\frac{1}{2} \mathrm{VI}$
3 Zero
4 $\mathrm{VI}^{2}$
Explanation:
C The average power dissipated is $\mathrm{P}_{\mathrm{avg}}=\mathrm{VI} \cos \phi$ Where, $\cos \phi$ is the power factor. For pure inductor, $\phi=\frac{\pi}{2}$ $\therefore \operatorname{Cos} \phi$ or $\mathrm{P}_{\text {avg }}=0$ Hence, there is no power dissipation in pure inductor.
155345
In an A.C. circuit, the current flowing in inductance is $I=5 \sin (100 t-\pi / 2)$ amperes and the potential difference is $V=200 \sin (100 t)$ volts. The power consumption is equal to
155346
In a $\mathrm{AC}$ circuit the voltage and current are described by $V=200 \sin \left(319 t-\frac{\pi}{6}\right)$ volts and $I=50 \sin \left(314 t+\frac{\pi}{6}\right) \mathrm{mA}$ respectively. The average power dissipated in the circuit is:
1 2.5 watts
2 5.0 watts
3 10.0 watts
4 50.0 watts
Explanation:
A Given, $\mathrm{V}=200 \sin \left(319 \mathrm{t}-\frac{\pi}{6}\right) \mathrm{V}$ $\mathrm{I}=50 \sin \left(314 \mathrm{t}+\frac{\pi}{6}\right) \mathrm{mA}$ On comparing equation, $\mathrm{V}=\mathrm{V}_{0} \sin (\omega \mathrm{t}+\phi)$ and $\mathrm{I}=\mathrm{I}_{0} \sin$ $(\omega \mathrm{t}+\phi)$ Then, $\mathrm{V}_{0}=200 \mathrm{~V}, \mathrm{I}_{0}=50 \mathrm{~mA}=50 \times 10^{-3} \mathrm{~A}$ $\phi=\phi_{1}-\phi_{2}=\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)=\frac{\pi}{3}$ The average power dissipated in the circuit - $\mathrm{P}_{\mathrm{avg}} =\frac{\mathrm{V}_{0} \mathrm{I}_{0}}{2} \cos \phi$ $\mathrm{P}_{\mathrm{avg}} =\frac{200 \times 50 \times 10^{-3}}{2} \cos 60^{\circ}$ $\mathrm{P}_{\mathrm{avg}} =2.5 \mathrm{~W}$
AIIMS-2011
Alternating Current
155347
An electric power station transmits $100 \mathrm{MW}$ power through long and thin cable. If the transmission is at (i) $20000 \mathrm{~V}$, (ii) $200 \mathrm{~V}$, in which case would be less power loss?
1 In (i) only
2 In (ii) only
3 In each case, power loss is zero
4 Data is insufficient
Explanation:
A Given, power of electric station $\mathrm{P}=100 \mathrm{MW}=100 \times 10^{6} \mathrm{~W}=10^{8} \mathrm{~W}$ (i) Given, $\mathrm{V}_{1}=20,000 \mathrm{~V}$ $I_{1}=\frac{P}{V_{1}}=\frac{10^{8}}{20,000}=\frac{10^{8}}{2 \times 10^{4}}=\frac{10000}{2}$ $I_{1}=5000 \mathrm{~A}$ Then rate of heat dissipation, $\mathrm{P}_{1}=\mathrm{I}_{1}^{2} \mathrm{R}$ $\mathrm{P}_{1}=(5000)^{2} \mathrm{R}$ $\mathrm{P}_{1}=25 \times 10^{6} \mathrm{R} \mathrm{W}$ (ii) Given, $\mathrm{V}_{2}=200 \mathrm{~V}$ $\mathrm{I}_{2}=\frac{\mathrm{P}}{\mathrm{V}_{2}}=\frac{10^{8}}{200}=5 \times 10^{5} \mathrm{~A}$ $\mathrm{P}_{2}=\mathrm{I}_{2}^{2} \mathrm{R}=\left(5 \times 10^{5}\right)^{2} \mathrm{R}$ $\mathrm{P}_{2}=25 \times 10^{10} \mathrm{RW}$ Comparing equation (i) and equation (ii), we get - There will be lesser power wastage when power is transmitted at 20,000 V
BCECE-2004
Alternating Current
155349
In an $\mathrm{AC}$ circuit, the instantaneous values of emf and current are $V=200 \sin (314 t) V$ and $I=\sin \left(314 t+\frac{\pi}{3}\right) A . \quad$ The average power consumed (in W) is
155350
The average power dissipated in a pure inductor is :
1 $\frac{\mathrm{VI}^{2}}{4}$
2 $\frac{1}{2} \mathrm{VI}$
3 Zero
4 $\mathrm{VI}^{2}$
Explanation:
C The average power dissipated is $\mathrm{P}_{\mathrm{avg}}=\mathrm{VI} \cos \phi$ Where, $\cos \phi$ is the power factor. For pure inductor, $\phi=\frac{\pi}{2}$ $\therefore \operatorname{Cos} \phi$ or $\mathrm{P}_{\text {avg }}=0$ Hence, there is no power dissipation in pure inductor.
