154972
A sinusoidal A.C. current flows through a resistor of resistance $10 \Omega$. If the peak current 2 A flowing through the resistor then the power dissipated in
1 30
2 20
3 10
4 40
Explanation:
B Given, Resistance, $\mathrm{R}=10 \Omega$ Peak current, $\mathrm{I}_{0}=2 \mathrm{~A}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \mathrm{Amp}$ We know that, The power dissipated, $\mathrm{P}=\mathrm{I}_{\text {rms }}^{2} \times \mathrm{R}=(\sqrt{2})^{2} \times 10=20 \mathrm{~W}$
GUJCET 2018
Alternating Current
154973
For the AC circuit shown below, phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414 \Omega$, then the values of $P$ and $Q$ are
1 $1 \mathrm{k} \Omega, 10 \mu \mathrm{F}$
2 $1 \mathrm{k} \Omega, 1 \mu \mathrm{F}$
3 $1 \mathrm{k} \Omega, 10 \mathrm{mH}$
4 $1 \mathrm{k} \Omega, 1 \mathrm{mH}$
Explanation:
A Given that, Impedance of the circuit $(Z)=1414 \Omega$ Difference between emf and currents $(\phi)=\frac{\pi}{4}$ In above figure, current is ahead of voltage, so its RC circuit. So, P is Resistance and Q is capacitor. Now RC circuit is Impedence, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{C}}^{2}}$ $Z=1414=1.414 \times 1000$ $Z=1000 \sqrt{2}$ So, $\quad Z=\sqrt{(1000)^{2}+(1000)^{2}}$ We compare with equation (i) $\mathrm{R}=1000 \Omega=1 \mathrm{k} \Omega, \mathrm{X}_{\mathrm{C}}=1000$ Now, $\text { Capacitor } \mathrm{C}=\frac{1}{\omega \mathrm{X}_{\mathrm{C}}}$ $=\frac{1}{100 \times 1000} \quad[\omega=100]$ $=10 \times 10^{-6} \mathrm{~F}=10 \mu \mathrm{F}$ The value of $\mathrm{P}$ and $\mathrm{Q}$ is $1 \mathrm{k} \Omega$ and $10 \mu \mathrm{F}$.
AMU-2012]
Alternating Current
154974
Voltage and current in an AC circuit are given by $V=5 \sin \left(100 \pi t-\frac{\pi}{6}\right) \text { and } I=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$
1 voltage leads the current by $30^{\circ}$
2 current leads the voltage by $30^{\circ}$
3 current leads the voltage by $60^{\circ}$
4 voltage leads the current by $60^{\circ}$
Explanation:
C Given that, Voltage of AC circuit $(V)=5 \sin \left(100 \pi t-\frac{\pi}{6}\right)$ Current of AC circuit (I) $=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$ The phase difference of AC current and voltage $\Delta \phi=\phi_{\mathrm{i}}-\phi_{\mathrm{v}}$ $=\left(100 \pi \mathrm{t}+\frac{\pi}{6}\right)-\left(100 \pi \mathrm{t}-\frac{\pi}{6}\right)$ $=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}=60^{\circ}$ Hence, current leads the voltage by $60^{\circ}$.
Manipal UGET-2018
Alternating Current
154975
In the A.C. circuit shown, keeping ' $K$ ' pressed, if an iron rod inserted into the coil, the bulb in the circuit
1 glows less brightly
2 glows with same brightness (as before the rod is inserted)
3 gets damaged
4 glows more brightly
Explanation:
A On inserting iron rod, the inductance $\left(\mathrm{X}_{\mathrm{L}}\right)$ of the coil increases $\text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}$ If the inductance increases then current will decrease and bulb will glows less brightly.
Karnataka CET-2017
Alternating Current
154978
An alternating voltage $E=200 \sqrt{2} \sin (100 t) \mathrm{V}$ is connected to a $1 \mu \mathrm{F}$ capacitor through an $\mathrm{AC}$ ammeter. The reading of ammeter is
154972
A sinusoidal A.C. current flows through a resistor of resistance $10 \Omega$. If the peak current 2 A flowing through the resistor then the power dissipated in
1 30
2 20
3 10
4 40
Explanation:
B Given, Resistance, $\mathrm{R}=10 \Omega$ Peak current, $\mathrm{I}_{0}=2 \mathrm{~A}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \mathrm{Amp}$ We know that, The power dissipated, $\mathrm{P}=\mathrm{I}_{\text {rms }}^{2} \times \mathrm{R}=(\sqrt{2})^{2} \times 10=20 \mathrm{~W}$
GUJCET 2018
Alternating Current
154973
For the AC circuit shown below, phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414 \Omega$, then the values of $P$ and $Q$ are
1 $1 \mathrm{k} \Omega, 10 \mu \mathrm{F}$
2 $1 \mathrm{k} \Omega, 1 \mu \mathrm{F}$
3 $1 \mathrm{k} \Omega, 10 \mathrm{mH}$
4 $1 \mathrm{k} \Omega, 1 \mathrm{mH}$
Explanation:
A Given that, Impedance of the circuit $(Z)=1414 \Omega$ Difference between emf and currents $(\phi)=\frac{\pi}{4}$ In above figure, current is ahead of voltage, so its RC circuit. So, P is Resistance and Q is capacitor. Now RC circuit is Impedence, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{C}}^{2}}$ $Z=1414=1.414 \times 1000$ $Z=1000 \sqrt{2}$ So, $\quad Z=\sqrt{(1000)^{2}+(1000)^{2}}$ We compare with equation (i) $\mathrm{R}=1000 \Omega=1 \mathrm{k} \Omega, \mathrm{X}_{\mathrm{C}}=1000$ Now, $\text { Capacitor } \mathrm{C}=\frac{1}{\omega \mathrm{X}_{\mathrm{C}}}$ $=\frac{1}{100 \times 1000} \quad[\omega=100]$ $=10 \times 10^{-6} \mathrm{~F}=10 \mu \mathrm{F}$ The value of $\mathrm{P}$ and $\mathrm{Q}$ is $1 \mathrm{k} \Omega$ and $10 \mu \mathrm{F}$.
AMU-2012]
Alternating Current
154974
Voltage and current in an AC circuit are given by $V=5 \sin \left(100 \pi t-\frac{\pi}{6}\right) \text { and } I=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$
1 voltage leads the current by $30^{\circ}$
2 current leads the voltage by $30^{\circ}$
3 current leads the voltage by $60^{\circ}$
4 voltage leads the current by $60^{\circ}$
Explanation:
C Given that, Voltage of AC circuit $(V)=5 \sin \left(100 \pi t-\frac{\pi}{6}\right)$ Current of AC circuit (I) $=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$ The phase difference of AC current and voltage $\Delta \phi=\phi_{\mathrm{i}}-\phi_{\mathrm{v}}$ $=\left(100 \pi \mathrm{t}+\frac{\pi}{6}\right)-\left(100 \pi \mathrm{t}-\frac{\pi}{6}\right)$ $=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}=60^{\circ}$ Hence, current leads the voltage by $60^{\circ}$.
Manipal UGET-2018
Alternating Current
154975
In the A.C. circuit shown, keeping ' $K$ ' pressed, if an iron rod inserted into the coil, the bulb in the circuit
1 glows less brightly
2 glows with same brightness (as before the rod is inserted)
3 gets damaged
4 glows more brightly
Explanation:
A On inserting iron rod, the inductance $\left(\mathrm{X}_{\mathrm{L}}\right)$ of the coil increases $\text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}$ If the inductance increases then current will decrease and bulb will glows less brightly.
Karnataka CET-2017
Alternating Current
154978
An alternating voltage $E=200 \sqrt{2} \sin (100 t) \mathrm{V}$ is connected to a $1 \mu \mathrm{F}$ capacitor through an $\mathrm{AC}$ ammeter. The reading of ammeter is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
154972
A sinusoidal A.C. current flows through a resistor of resistance $10 \Omega$. If the peak current 2 A flowing through the resistor then the power dissipated in
1 30
2 20
3 10
4 40
Explanation:
B Given, Resistance, $\mathrm{R}=10 \Omega$ Peak current, $\mathrm{I}_{0}=2 \mathrm{~A}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \mathrm{Amp}$ We know that, The power dissipated, $\mathrm{P}=\mathrm{I}_{\text {rms }}^{2} \times \mathrm{R}=(\sqrt{2})^{2} \times 10=20 \mathrm{~W}$
GUJCET 2018
Alternating Current
154973
For the AC circuit shown below, phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414 \Omega$, then the values of $P$ and $Q$ are
1 $1 \mathrm{k} \Omega, 10 \mu \mathrm{F}$
2 $1 \mathrm{k} \Omega, 1 \mu \mathrm{F}$
3 $1 \mathrm{k} \Omega, 10 \mathrm{mH}$
4 $1 \mathrm{k} \Omega, 1 \mathrm{mH}$
Explanation:
A Given that, Impedance of the circuit $(Z)=1414 \Omega$ Difference between emf and currents $(\phi)=\frac{\pi}{4}$ In above figure, current is ahead of voltage, so its RC circuit. So, P is Resistance and Q is capacitor. Now RC circuit is Impedence, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{C}}^{2}}$ $Z=1414=1.414 \times 1000$ $Z=1000 \sqrt{2}$ So, $\quad Z=\sqrt{(1000)^{2}+(1000)^{2}}$ We compare with equation (i) $\mathrm{R}=1000 \Omega=1 \mathrm{k} \Omega, \mathrm{X}_{\mathrm{C}}=1000$ Now, $\text { Capacitor } \mathrm{C}=\frac{1}{\omega \mathrm{X}_{\mathrm{C}}}$ $=\frac{1}{100 \times 1000} \quad[\omega=100]$ $=10 \times 10^{-6} \mathrm{~F}=10 \mu \mathrm{F}$ The value of $\mathrm{P}$ and $\mathrm{Q}$ is $1 \mathrm{k} \Omega$ and $10 \mu \mathrm{F}$.
AMU-2012]
Alternating Current
154974
Voltage and current in an AC circuit are given by $V=5 \sin \left(100 \pi t-\frac{\pi}{6}\right) \text { and } I=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$
1 voltage leads the current by $30^{\circ}$
2 current leads the voltage by $30^{\circ}$
3 current leads the voltage by $60^{\circ}$
4 voltage leads the current by $60^{\circ}$
Explanation:
C Given that, Voltage of AC circuit $(V)=5 \sin \left(100 \pi t-\frac{\pi}{6}\right)$ Current of AC circuit (I) $=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$ The phase difference of AC current and voltage $\Delta \phi=\phi_{\mathrm{i}}-\phi_{\mathrm{v}}$ $=\left(100 \pi \mathrm{t}+\frac{\pi}{6}\right)-\left(100 \pi \mathrm{t}-\frac{\pi}{6}\right)$ $=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}=60^{\circ}$ Hence, current leads the voltage by $60^{\circ}$.
Manipal UGET-2018
Alternating Current
154975
In the A.C. circuit shown, keeping ' $K$ ' pressed, if an iron rod inserted into the coil, the bulb in the circuit
1 glows less brightly
2 glows with same brightness (as before the rod is inserted)
3 gets damaged
4 glows more brightly
Explanation:
A On inserting iron rod, the inductance $\left(\mathrm{X}_{\mathrm{L}}\right)$ of the coil increases $\text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}$ If the inductance increases then current will decrease and bulb will glows less brightly.
Karnataka CET-2017
Alternating Current
154978
An alternating voltage $E=200 \sqrt{2} \sin (100 t) \mathrm{V}$ is connected to a $1 \mu \mathrm{F}$ capacitor through an $\mathrm{AC}$ ammeter. The reading of ammeter is
154972
A sinusoidal A.C. current flows through a resistor of resistance $10 \Omega$. If the peak current 2 A flowing through the resistor then the power dissipated in
1 30
2 20
3 10
4 40
Explanation:
B Given, Resistance, $\mathrm{R}=10 \Omega$ Peak current, $\mathrm{I}_{0}=2 \mathrm{~A}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \mathrm{Amp}$ We know that, The power dissipated, $\mathrm{P}=\mathrm{I}_{\text {rms }}^{2} \times \mathrm{R}=(\sqrt{2})^{2} \times 10=20 \mathrm{~W}$
GUJCET 2018
Alternating Current
154973
For the AC circuit shown below, phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414 \Omega$, then the values of $P$ and $Q$ are
1 $1 \mathrm{k} \Omega, 10 \mu \mathrm{F}$
2 $1 \mathrm{k} \Omega, 1 \mu \mathrm{F}$
3 $1 \mathrm{k} \Omega, 10 \mathrm{mH}$
4 $1 \mathrm{k} \Omega, 1 \mathrm{mH}$
Explanation:
A Given that, Impedance of the circuit $(Z)=1414 \Omega$ Difference between emf and currents $(\phi)=\frac{\pi}{4}$ In above figure, current is ahead of voltage, so its RC circuit. So, P is Resistance and Q is capacitor. Now RC circuit is Impedence, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{C}}^{2}}$ $Z=1414=1.414 \times 1000$ $Z=1000 \sqrt{2}$ So, $\quad Z=\sqrt{(1000)^{2}+(1000)^{2}}$ We compare with equation (i) $\mathrm{R}=1000 \Omega=1 \mathrm{k} \Omega, \mathrm{X}_{\mathrm{C}}=1000$ Now, $\text { Capacitor } \mathrm{C}=\frac{1}{\omega \mathrm{X}_{\mathrm{C}}}$ $=\frac{1}{100 \times 1000} \quad[\omega=100]$ $=10 \times 10^{-6} \mathrm{~F}=10 \mu \mathrm{F}$ The value of $\mathrm{P}$ and $\mathrm{Q}$ is $1 \mathrm{k} \Omega$ and $10 \mu \mathrm{F}$.
AMU-2012]
Alternating Current
154974
Voltage and current in an AC circuit are given by $V=5 \sin \left(100 \pi t-\frac{\pi}{6}\right) \text { and } I=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$
1 voltage leads the current by $30^{\circ}$
2 current leads the voltage by $30^{\circ}$
3 current leads the voltage by $60^{\circ}$
4 voltage leads the current by $60^{\circ}$
Explanation:
C Given that, Voltage of AC circuit $(V)=5 \sin \left(100 \pi t-\frac{\pi}{6}\right)$ Current of AC circuit (I) $=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$ The phase difference of AC current and voltage $\Delta \phi=\phi_{\mathrm{i}}-\phi_{\mathrm{v}}$ $=\left(100 \pi \mathrm{t}+\frac{\pi}{6}\right)-\left(100 \pi \mathrm{t}-\frac{\pi}{6}\right)$ $=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}=60^{\circ}$ Hence, current leads the voltage by $60^{\circ}$.
Manipal UGET-2018
Alternating Current
154975
In the A.C. circuit shown, keeping ' $K$ ' pressed, if an iron rod inserted into the coil, the bulb in the circuit
1 glows less brightly
2 glows with same brightness (as before the rod is inserted)
3 gets damaged
4 glows more brightly
Explanation:
A On inserting iron rod, the inductance $\left(\mathrm{X}_{\mathrm{L}}\right)$ of the coil increases $\text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}$ If the inductance increases then current will decrease and bulb will glows less brightly.
Karnataka CET-2017
Alternating Current
154978
An alternating voltage $E=200 \sqrt{2} \sin (100 t) \mathrm{V}$ is connected to a $1 \mu \mathrm{F}$ capacitor through an $\mathrm{AC}$ ammeter. The reading of ammeter is
154972
A sinusoidal A.C. current flows through a resistor of resistance $10 \Omega$. If the peak current 2 A flowing through the resistor then the power dissipated in
1 30
2 20
3 10
4 40
Explanation:
B Given, Resistance, $\mathrm{R}=10 \Omega$ Peak current, $\mathrm{I}_{0}=2 \mathrm{~A}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \mathrm{Amp}$ We know that, The power dissipated, $\mathrm{P}=\mathrm{I}_{\text {rms }}^{2} \times \mathrm{R}=(\sqrt{2})^{2} \times 10=20 \mathrm{~W}$
GUJCET 2018
Alternating Current
154973
For the AC circuit shown below, phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414 \Omega$, then the values of $P$ and $Q$ are
1 $1 \mathrm{k} \Omega, 10 \mu \mathrm{F}$
2 $1 \mathrm{k} \Omega, 1 \mu \mathrm{F}$
3 $1 \mathrm{k} \Omega, 10 \mathrm{mH}$
4 $1 \mathrm{k} \Omega, 1 \mathrm{mH}$
Explanation:
A Given that, Impedance of the circuit $(Z)=1414 \Omega$ Difference between emf and currents $(\phi)=\frac{\pi}{4}$ In above figure, current is ahead of voltage, so its RC circuit. So, P is Resistance and Q is capacitor. Now RC circuit is Impedence, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{C}}^{2}}$ $Z=1414=1.414 \times 1000$ $Z=1000 \sqrt{2}$ So, $\quad Z=\sqrt{(1000)^{2}+(1000)^{2}}$ We compare with equation (i) $\mathrm{R}=1000 \Omega=1 \mathrm{k} \Omega, \mathrm{X}_{\mathrm{C}}=1000$ Now, $\text { Capacitor } \mathrm{C}=\frac{1}{\omega \mathrm{X}_{\mathrm{C}}}$ $=\frac{1}{100 \times 1000} \quad[\omega=100]$ $=10 \times 10^{-6} \mathrm{~F}=10 \mu \mathrm{F}$ The value of $\mathrm{P}$ and $\mathrm{Q}$ is $1 \mathrm{k} \Omega$ and $10 \mu \mathrm{F}$.
AMU-2012]
Alternating Current
154974
Voltage and current in an AC circuit are given by $V=5 \sin \left(100 \pi t-\frac{\pi}{6}\right) \text { and } I=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$
1 voltage leads the current by $30^{\circ}$
2 current leads the voltage by $30^{\circ}$
3 current leads the voltage by $60^{\circ}$
4 voltage leads the current by $60^{\circ}$
Explanation:
C Given that, Voltage of AC circuit $(V)=5 \sin \left(100 \pi t-\frac{\pi}{6}\right)$ Current of AC circuit (I) $=4 \sin \left(100 \pi t+\frac{\pi}{6}\right)$ The phase difference of AC current and voltage $\Delta \phi=\phi_{\mathrm{i}}-\phi_{\mathrm{v}}$ $=\left(100 \pi \mathrm{t}+\frac{\pi}{6}\right)-\left(100 \pi \mathrm{t}-\frac{\pi}{6}\right)$ $=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}=60^{\circ}$ Hence, current leads the voltage by $60^{\circ}$.
Manipal UGET-2018
Alternating Current
154975
In the A.C. circuit shown, keeping ' $K$ ' pressed, if an iron rod inserted into the coil, the bulb in the circuit
1 glows less brightly
2 glows with same brightness (as before the rod is inserted)
3 gets damaged
4 glows more brightly
Explanation:
A On inserting iron rod, the inductance $\left(\mathrm{X}_{\mathrm{L}}\right)$ of the coil increases $\text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}$ If the inductance increases then current will decrease and bulb will glows less brightly.
Karnataka CET-2017
Alternating Current
154978
An alternating voltage $E=200 \sqrt{2} \sin (100 t) \mathrm{V}$ is connected to a $1 \mu \mathrm{F}$ capacitor through an $\mathrm{AC}$ ammeter. The reading of ammeter is
