NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155007
Using an $\mathrm{AC}$ voltmeter the potential difference in the electrical line in a house is read to be $234 \mathrm{~V}$. If the line frequency is known to be 50 cycles/second, the equation for the line voltage is given as
1 $\mathrm{V}=165 \sin (100 \pi \mathrm{t})$
2 $V=331 \sin (100 \pi t)$
3 $\mathrm{V}=220 \sin (100 \pi \mathrm{t})$
4 $V=440 \sin (100 \pi t)$
Explanation:
B Given, $\mathrm{V}_{\mathrm{rms}}=234 \mathrm{~V}$ Frequency, $\mathrm{f}=50 \mathrm{cycle} / \mathrm{sec}$ Peak voltage, $\mathrm{V}_{0}=\sqrt{2} \mathrm{~V}_{\mathrm{rms}}=\sqrt{2} \times 234=331 \mathrm{~V}$ And $\omega=2 \pi \mathrm{ft}=2 \pi \times 50 \times \mathrm{t}=100 \pi \mathrm{t}$ Thus, the eq ${ }^{\mathrm{n}}$ of the line voltage is, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}=331 \sin (100 \pi \mathrm{t})$
VITEEE-2010
Alternating Current
155010
An ideal resistance $R$, ideal inductance $L$, ideal capacitance $C$ and $A C$ voltmeters $V_{1}, V_{2}, V_{3}$ and $V_{4}$ are connected to an $A C$ source as shown. At resonance,
1 reading in $\mathrm{V}_{2}=$ reading in $\mathrm{V}_{3}$
2 reading in $\mathrm{V}_{3}=$ reading in $\mathrm{V}_{1}$
3 reading in $V_{1}=$ reading in $V_{2}$
4 reading in $\mathrm{V}_{2}=$ reading in $\mathrm{V}_{4}$
Explanation:
A In RLC circuit At resonance condition, Voltage across inductor $=$ Voltage across capacitor $\mathrm{V}_{\mathrm{L}}=\mathrm{V}_{\mathrm{C}}$ Hence, reading in $V_{2}=$ reading in $V_{3}$
Karnataka CET-2012
Alternating Current
155014
If reading of an ammeter is $10 \mathrm{~A}$, the peak value of current is
1 $\frac{10}{\sqrt{2}} \mathrm{~A}$
2 $\frac{5}{\sqrt{2}} \mathrm{~A}$
3 $20 \sqrt{2} \mathrm{~A}$
4 $10 \sqrt{2} \mathrm{~A}$
Explanation:
D Given, $\mathrm{I}_{\mathrm{rms}}=10 \mathrm{~A}$ We know that, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}$ Alternating Current $\mathrm{I}_{0}=\sqrt{2} \times \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times 10=10 \sqrt{2} \mathrm{~A}$ Hence, the peak value of current, $I_{0}=10 \sqrt{2} \mathrm{~A}$.
UP CPMT-2003
Alternating Current
155016
Average power generated in an inductor connected to an $\mathrm{AC}$ source is
1 $\frac{1}{2} \mathrm{Li}^{2}$
2 $\mathrm{Li}^{2}$
3 zero
4 none of these
Explanation:
C When an AC source is connected to an inductor the phase difference between current and applied voltage $\phi=\pi / 2$ Average power $(\mathrm{P})=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}_{\text {avg }}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos (\pi / 2)$ $\mathrm{P}_{\mathrm{avg}}=0$ So, the power generated in an inductor connected to an AC source is zero.
155007
Using an $\mathrm{AC}$ voltmeter the potential difference in the electrical line in a house is read to be $234 \mathrm{~V}$. If the line frequency is known to be 50 cycles/second, the equation for the line voltage is given as
1 $\mathrm{V}=165 \sin (100 \pi \mathrm{t})$
2 $V=331 \sin (100 \pi t)$
3 $\mathrm{V}=220 \sin (100 \pi \mathrm{t})$
4 $V=440 \sin (100 \pi t)$
Explanation:
B Given, $\mathrm{V}_{\mathrm{rms}}=234 \mathrm{~V}$ Frequency, $\mathrm{f}=50 \mathrm{cycle} / \mathrm{sec}$ Peak voltage, $\mathrm{V}_{0}=\sqrt{2} \mathrm{~V}_{\mathrm{rms}}=\sqrt{2} \times 234=331 \mathrm{~V}$ And $\omega=2 \pi \mathrm{ft}=2 \pi \times 50 \times \mathrm{t}=100 \pi \mathrm{t}$ Thus, the eq ${ }^{\mathrm{n}}$ of the line voltage is, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}=331 \sin (100 \pi \mathrm{t})$
VITEEE-2010
Alternating Current
155010
An ideal resistance $R$, ideal inductance $L$, ideal capacitance $C$ and $A C$ voltmeters $V_{1}, V_{2}, V_{3}$ and $V_{4}$ are connected to an $A C$ source as shown. At resonance,
1 reading in $\mathrm{V}_{2}=$ reading in $\mathrm{V}_{3}$
2 reading in $\mathrm{V}_{3}=$ reading in $\mathrm{V}_{1}$
3 reading in $V_{1}=$ reading in $V_{2}$
4 reading in $\mathrm{V}_{2}=$ reading in $\mathrm{V}_{4}$
Explanation:
A In RLC circuit At resonance condition, Voltage across inductor $=$ Voltage across capacitor $\mathrm{V}_{\mathrm{L}}=\mathrm{V}_{\mathrm{C}}$ Hence, reading in $V_{2}=$ reading in $V_{3}$
Karnataka CET-2012
Alternating Current
155014
If reading of an ammeter is $10 \mathrm{~A}$, the peak value of current is
1 $\frac{10}{\sqrt{2}} \mathrm{~A}$
2 $\frac{5}{\sqrt{2}} \mathrm{~A}$
3 $20 \sqrt{2} \mathrm{~A}$
4 $10 \sqrt{2} \mathrm{~A}$
Explanation:
D Given, $\mathrm{I}_{\mathrm{rms}}=10 \mathrm{~A}$ We know that, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}$ Alternating Current $\mathrm{I}_{0}=\sqrt{2} \times \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times 10=10 \sqrt{2} \mathrm{~A}$ Hence, the peak value of current, $I_{0}=10 \sqrt{2} \mathrm{~A}$.
UP CPMT-2003
Alternating Current
155016
Average power generated in an inductor connected to an $\mathrm{AC}$ source is
1 $\frac{1}{2} \mathrm{Li}^{2}$
2 $\mathrm{Li}^{2}$
3 zero
4 none of these
Explanation:
C When an AC source is connected to an inductor the phase difference between current and applied voltage $\phi=\pi / 2$ Average power $(\mathrm{P})=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}_{\text {avg }}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos (\pi / 2)$ $\mathrm{P}_{\mathrm{avg}}=0$ So, the power generated in an inductor connected to an AC source is zero.
155007
Using an $\mathrm{AC}$ voltmeter the potential difference in the electrical line in a house is read to be $234 \mathrm{~V}$. If the line frequency is known to be 50 cycles/second, the equation for the line voltage is given as
1 $\mathrm{V}=165 \sin (100 \pi \mathrm{t})$
2 $V=331 \sin (100 \pi t)$
3 $\mathrm{V}=220 \sin (100 \pi \mathrm{t})$
4 $V=440 \sin (100 \pi t)$
Explanation:
B Given, $\mathrm{V}_{\mathrm{rms}}=234 \mathrm{~V}$ Frequency, $\mathrm{f}=50 \mathrm{cycle} / \mathrm{sec}$ Peak voltage, $\mathrm{V}_{0}=\sqrt{2} \mathrm{~V}_{\mathrm{rms}}=\sqrt{2} \times 234=331 \mathrm{~V}$ And $\omega=2 \pi \mathrm{ft}=2 \pi \times 50 \times \mathrm{t}=100 \pi \mathrm{t}$ Thus, the eq ${ }^{\mathrm{n}}$ of the line voltage is, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}=331 \sin (100 \pi \mathrm{t})$
VITEEE-2010
Alternating Current
155010
An ideal resistance $R$, ideal inductance $L$, ideal capacitance $C$ and $A C$ voltmeters $V_{1}, V_{2}, V_{3}$ and $V_{4}$ are connected to an $A C$ source as shown. At resonance,
1 reading in $\mathrm{V}_{2}=$ reading in $\mathrm{V}_{3}$
2 reading in $\mathrm{V}_{3}=$ reading in $\mathrm{V}_{1}$
3 reading in $V_{1}=$ reading in $V_{2}$
4 reading in $\mathrm{V}_{2}=$ reading in $\mathrm{V}_{4}$
Explanation:
A In RLC circuit At resonance condition, Voltage across inductor $=$ Voltage across capacitor $\mathrm{V}_{\mathrm{L}}=\mathrm{V}_{\mathrm{C}}$ Hence, reading in $V_{2}=$ reading in $V_{3}$
Karnataka CET-2012
Alternating Current
155014
If reading of an ammeter is $10 \mathrm{~A}$, the peak value of current is
1 $\frac{10}{\sqrt{2}} \mathrm{~A}$
2 $\frac{5}{\sqrt{2}} \mathrm{~A}$
3 $20 \sqrt{2} \mathrm{~A}$
4 $10 \sqrt{2} \mathrm{~A}$
Explanation:
D Given, $\mathrm{I}_{\mathrm{rms}}=10 \mathrm{~A}$ We know that, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}$ Alternating Current $\mathrm{I}_{0}=\sqrt{2} \times \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times 10=10 \sqrt{2} \mathrm{~A}$ Hence, the peak value of current, $I_{0}=10 \sqrt{2} \mathrm{~A}$.
UP CPMT-2003
Alternating Current
155016
Average power generated in an inductor connected to an $\mathrm{AC}$ source is
1 $\frac{1}{2} \mathrm{Li}^{2}$
2 $\mathrm{Li}^{2}$
3 zero
4 none of these
Explanation:
C When an AC source is connected to an inductor the phase difference between current and applied voltage $\phi=\pi / 2$ Average power $(\mathrm{P})=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}_{\text {avg }}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos (\pi / 2)$ $\mathrm{P}_{\mathrm{avg}}=0$ So, the power generated in an inductor connected to an AC source is zero.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155007
Using an $\mathrm{AC}$ voltmeter the potential difference in the electrical line in a house is read to be $234 \mathrm{~V}$. If the line frequency is known to be 50 cycles/second, the equation for the line voltage is given as
1 $\mathrm{V}=165 \sin (100 \pi \mathrm{t})$
2 $V=331 \sin (100 \pi t)$
3 $\mathrm{V}=220 \sin (100 \pi \mathrm{t})$
4 $V=440 \sin (100 \pi t)$
Explanation:
B Given, $\mathrm{V}_{\mathrm{rms}}=234 \mathrm{~V}$ Frequency, $\mathrm{f}=50 \mathrm{cycle} / \mathrm{sec}$ Peak voltage, $\mathrm{V}_{0}=\sqrt{2} \mathrm{~V}_{\mathrm{rms}}=\sqrt{2} \times 234=331 \mathrm{~V}$ And $\omega=2 \pi \mathrm{ft}=2 \pi \times 50 \times \mathrm{t}=100 \pi \mathrm{t}$ Thus, the eq ${ }^{\mathrm{n}}$ of the line voltage is, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}=331 \sin (100 \pi \mathrm{t})$
VITEEE-2010
Alternating Current
155010
An ideal resistance $R$, ideal inductance $L$, ideal capacitance $C$ and $A C$ voltmeters $V_{1}, V_{2}, V_{3}$ and $V_{4}$ are connected to an $A C$ source as shown. At resonance,
1 reading in $\mathrm{V}_{2}=$ reading in $\mathrm{V}_{3}$
2 reading in $\mathrm{V}_{3}=$ reading in $\mathrm{V}_{1}$
3 reading in $V_{1}=$ reading in $V_{2}$
4 reading in $\mathrm{V}_{2}=$ reading in $\mathrm{V}_{4}$
Explanation:
A In RLC circuit At resonance condition, Voltage across inductor $=$ Voltage across capacitor $\mathrm{V}_{\mathrm{L}}=\mathrm{V}_{\mathrm{C}}$ Hence, reading in $V_{2}=$ reading in $V_{3}$
Karnataka CET-2012
Alternating Current
155014
If reading of an ammeter is $10 \mathrm{~A}$, the peak value of current is
1 $\frac{10}{\sqrt{2}} \mathrm{~A}$
2 $\frac{5}{\sqrt{2}} \mathrm{~A}$
3 $20 \sqrt{2} \mathrm{~A}$
4 $10 \sqrt{2} \mathrm{~A}$
Explanation:
D Given, $\mathrm{I}_{\mathrm{rms}}=10 \mathrm{~A}$ We know that, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}$ Alternating Current $\mathrm{I}_{0}=\sqrt{2} \times \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times 10=10 \sqrt{2} \mathrm{~A}$ Hence, the peak value of current, $I_{0}=10 \sqrt{2} \mathrm{~A}$.
UP CPMT-2003
Alternating Current
155016
Average power generated in an inductor connected to an $\mathrm{AC}$ source is
1 $\frac{1}{2} \mathrm{Li}^{2}$
2 $\mathrm{Li}^{2}$
3 zero
4 none of these
Explanation:
C When an AC source is connected to an inductor the phase difference between current and applied voltage $\phi=\pi / 2$ Average power $(\mathrm{P})=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}_{\text {avg }}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos (\pi / 2)$ $\mathrm{P}_{\mathrm{avg}}=0$ So, the power generated in an inductor connected to an AC source is zero.
