154979
A $10 \mu \mathrm{F}$ capacitor is connected across a $200 \mathrm{~V}$, $50 \mathrm{~Hz}$ AC supply. The peak current through the circuit is :
1 $0.6 \sqrt{2} \mathrm{~A}$
2 $0.6 \mathrm{~A}$
3 $\frac{0.6 \pi}{2} \mathrm{~A}$
4 $\frac{0.6}{\sqrt{2}} \mathrm{~A}$
Explanation:
A Given, Capacitor, $\mathrm{C}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}$ Voltage,$\mathrm{V}=200 \mathrm{~V}$ Frequency of AC supply $=50 \mathrm{~Hz}$ We know that, impedance of a capacitor is - $X_{C}=\frac{1}{\omega C}, \quad X_{C}=\frac{1}{2 \pi f C}$ $X_{C}=\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}$ Rms current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}$ $I=\frac{200}{\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}}$ $I=200 \times 2 \pi \times 50 \times 10 \times 10^{-6}$ $I=0.628 \mathrm{~A}$ Peak current, $\mathrm{I}_{\max }=\sqrt{2} \mathrm{I}=\sqrt{2} \times 0.628$ $=0.628 \sqrt{2} \mathrm{~A} \approx 0.6 \sqrt{2} \mathrm{~A}$
JCECE-2004
Alternating Current
154980
The alternating voltage and current in an electric circuit are respectively given by $E=100 \sin 100 \pi \mathrm{t}, \mathrm{I}=5 \sin 100 \pi \mathrm{t}$ The reactance of the circuit will be
1 $1 \Omega$
2 $0.05 \Omega$
3 $20 \Omega$
4 zero
Explanation:
C Given that, Alternating voltage, $\mathrm{E}=100 \sin 100 \pi \mathrm{t}$ Current, $\mathrm{I}=5 \sin 100 \pi \mathrm{t}$ From equation (i) \& (ii), we get- $\mathrm{E}_{0}=100 \mathrm{~V}$ $\mathrm{I}_{0}=5 \mathrm{~A}$ $\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~V}$ And, $\quad I_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{5}{\sqrt{2}}$ Now, we know that- $\text { Reactance of circuit }(\mathrm{R}) =\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{I}_{\mathrm{rms}}}$ $\mathrm{R} =\frac{100 / \sqrt{2}}{5 / \sqrt{2}}=20 \Omega$
Manipal UGET-2015
Alternating Current
154981
The peak value of an alternating emf $E$ given by $E=E_{0} \cos \omega t$ is $10 \mathrm{~V}$ and its frequency is 50 Hz. At a time $t=\left(\frac{1}{600}\right) s$, the instantaneous value of the emf is
1 $10 \mathrm{~V}$
2 $5 \sqrt{3} \mathrm{~V}$
3 $5 \mathrm{~V}$
4 $1 \mathrm{~V}$
Explanation:
B Given that, Voltage, $\mathrm{E}_{0}=10 \mathrm{~V}$ Frequency $(\mathrm{f})=50 \mathrm{~Hz}$ Time $(\mathrm{t})=\frac{1}{600} \mathrm{~s}$ Now, instantaneous value of emf, $E=E_{0} \cos \omega \mathrm{t}=E_{0} \cos (2 \pi f t)$ $E=10 \cos \left(2 \pi \times 50 \times \frac{1}{600}\right) \quad[\omega=2 \pi f]$ $E=10 \cos \left(\frac{\pi}{6}\right)$ $E=10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3} V$
Manipal UGET-2010
Alternating Current
154982
On an $\mathrm{AC}$ circuit, the hot wire ammeter reads current $10 \mathrm{~A}$. Its peak value is
1 $10 \mathrm{~A}$
2 $20 \mathrm{~A}$
3 $14.14 \mathrm{~A}$
4 $7.07 \mathrm{~A}$
Explanation:
C The ammeter reads the rms value of current. So, $\quad \mathrm{I}_{\mathrm{rms}}=10 \mathrm{~A}$ $\\ \text { Irms } =\frac{\mathrm{I}_{\text {peak }}}{\sqrt{2}}$ $\mathrm{I}_{\text {peak }} =\sqrt{2} \mathrm{I}_{\text {rms }}$ $\mathrm{I}_{\text {peak }} =1.414 \times 10$ $\mathrm{I}_{\text {peak }} =14.14 \mathrm{~A}$
154979
A $10 \mu \mathrm{F}$ capacitor is connected across a $200 \mathrm{~V}$, $50 \mathrm{~Hz}$ AC supply. The peak current through the circuit is :
1 $0.6 \sqrt{2} \mathrm{~A}$
2 $0.6 \mathrm{~A}$
3 $\frac{0.6 \pi}{2} \mathrm{~A}$
4 $\frac{0.6}{\sqrt{2}} \mathrm{~A}$
Explanation:
A Given, Capacitor, $\mathrm{C}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}$ Voltage,$\mathrm{V}=200 \mathrm{~V}$ Frequency of AC supply $=50 \mathrm{~Hz}$ We know that, impedance of a capacitor is - $X_{C}=\frac{1}{\omega C}, \quad X_{C}=\frac{1}{2 \pi f C}$ $X_{C}=\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}$ Rms current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}$ $I=\frac{200}{\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}}$ $I=200 \times 2 \pi \times 50 \times 10 \times 10^{-6}$ $I=0.628 \mathrm{~A}$ Peak current, $\mathrm{I}_{\max }=\sqrt{2} \mathrm{I}=\sqrt{2} \times 0.628$ $=0.628 \sqrt{2} \mathrm{~A} \approx 0.6 \sqrt{2} \mathrm{~A}$
JCECE-2004
Alternating Current
154980
The alternating voltage and current in an electric circuit are respectively given by $E=100 \sin 100 \pi \mathrm{t}, \mathrm{I}=5 \sin 100 \pi \mathrm{t}$ The reactance of the circuit will be
1 $1 \Omega$
2 $0.05 \Omega$
3 $20 \Omega$
4 zero
Explanation:
C Given that, Alternating voltage, $\mathrm{E}=100 \sin 100 \pi \mathrm{t}$ Current, $\mathrm{I}=5 \sin 100 \pi \mathrm{t}$ From equation (i) \& (ii), we get- $\mathrm{E}_{0}=100 \mathrm{~V}$ $\mathrm{I}_{0}=5 \mathrm{~A}$ $\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~V}$ And, $\quad I_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{5}{\sqrt{2}}$ Now, we know that- $\text { Reactance of circuit }(\mathrm{R}) =\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{I}_{\mathrm{rms}}}$ $\mathrm{R} =\frac{100 / \sqrt{2}}{5 / \sqrt{2}}=20 \Omega$
Manipal UGET-2015
Alternating Current
154981
The peak value of an alternating emf $E$ given by $E=E_{0} \cos \omega t$ is $10 \mathrm{~V}$ and its frequency is 50 Hz. At a time $t=\left(\frac{1}{600}\right) s$, the instantaneous value of the emf is
1 $10 \mathrm{~V}$
2 $5 \sqrt{3} \mathrm{~V}$
3 $5 \mathrm{~V}$
4 $1 \mathrm{~V}$
Explanation:
B Given that, Voltage, $\mathrm{E}_{0}=10 \mathrm{~V}$ Frequency $(\mathrm{f})=50 \mathrm{~Hz}$ Time $(\mathrm{t})=\frac{1}{600} \mathrm{~s}$ Now, instantaneous value of emf, $E=E_{0} \cos \omega \mathrm{t}=E_{0} \cos (2 \pi f t)$ $E=10 \cos \left(2 \pi \times 50 \times \frac{1}{600}\right) \quad[\omega=2 \pi f]$ $E=10 \cos \left(\frac{\pi}{6}\right)$ $E=10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3} V$
Manipal UGET-2010
Alternating Current
154982
On an $\mathrm{AC}$ circuit, the hot wire ammeter reads current $10 \mathrm{~A}$. Its peak value is
1 $10 \mathrm{~A}$
2 $20 \mathrm{~A}$
3 $14.14 \mathrm{~A}$
4 $7.07 \mathrm{~A}$
Explanation:
C The ammeter reads the rms value of current. So, $\quad \mathrm{I}_{\mathrm{rms}}=10 \mathrm{~A}$ $\\ \text { Irms } =\frac{\mathrm{I}_{\text {peak }}}{\sqrt{2}}$ $\mathrm{I}_{\text {peak }} =\sqrt{2} \mathrm{I}_{\text {rms }}$ $\mathrm{I}_{\text {peak }} =1.414 \times 10$ $\mathrm{I}_{\text {peak }} =14.14 \mathrm{~A}$
154979
A $10 \mu \mathrm{F}$ capacitor is connected across a $200 \mathrm{~V}$, $50 \mathrm{~Hz}$ AC supply. The peak current through the circuit is :
1 $0.6 \sqrt{2} \mathrm{~A}$
2 $0.6 \mathrm{~A}$
3 $\frac{0.6 \pi}{2} \mathrm{~A}$
4 $\frac{0.6}{\sqrt{2}} \mathrm{~A}$
Explanation:
A Given, Capacitor, $\mathrm{C}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}$ Voltage,$\mathrm{V}=200 \mathrm{~V}$ Frequency of AC supply $=50 \mathrm{~Hz}$ We know that, impedance of a capacitor is - $X_{C}=\frac{1}{\omega C}, \quad X_{C}=\frac{1}{2 \pi f C}$ $X_{C}=\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}$ Rms current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}$ $I=\frac{200}{\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}}$ $I=200 \times 2 \pi \times 50 \times 10 \times 10^{-6}$ $I=0.628 \mathrm{~A}$ Peak current, $\mathrm{I}_{\max }=\sqrt{2} \mathrm{I}=\sqrt{2} \times 0.628$ $=0.628 \sqrt{2} \mathrm{~A} \approx 0.6 \sqrt{2} \mathrm{~A}$
JCECE-2004
Alternating Current
154980
The alternating voltage and current in an electric circuit are respectively given by $E=100 \sin 100 \pi \mathrm{t}, \mathrm{I}=5 \sin 100 \pi \mathrm{t}$ The reactance of the circuit will be
1 $1 \Omega$
2 $0.05 \Omega$
3 $20 \Omega$
4 zero
Explanation:
C Given that, Alternating voltage, $\mathrm{E}=100 \sin 100 \pi \mathrm{t}$ Current, $\mathrm{I}=5 \sin 100 \pi \mathrm{t}$ From equation (i) \& (ii), we get- $\mathrm{E}_{0}=100 \mathrm{~V}$ $\mathrm{I}_{0}=5 \mathrm{~A}$ $\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~V}$ And, $\quad I_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{5}{\sqrt{2}}$ Now, we know that- $\text { Reactance of circuit }(\mathrm{R}) =\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{I}_{\mathrm{rms}}}$ $\mathrm{R} =\frac{100 / \sqrt{2}}{5 / \sqrt{2}}=20 \Omega$
Manipal UGET-2015
Alternating Current
154981
The peak value of an alternating emf $E$ given by $E=E_{0} \cos \omega t$ is $10 \mathrm{~V}$ and its frequency is 50 Hz. At a time $t=\left(\frac{1}{600}\right) s$, the instantaneous value of the emf is
1 $10 \mathrm{~V}$
2 $5 \sqrt{3} \mathrm{~V}$
3 $5 \mathrm{~V}$
4 $1 \mathrm{~V}$
Explanation:
B Given that, Voltage, $\mathrm{E}_{0}=10 \mathrm{~V}$ Frequency $(\mathrm{f})=50 \mathrm{~Hz}$ Time $(\mathrm{t})=\frac{1}{600} \mathrm{~s}$ Now, instantaneous value of emf, $E=E_{0} \cos \omega \mathrm{t}=E_{0} \cos (2 \pi f t)$ $E=10 \cos \left(2 \pi \times 50 \times \frac{1}{600}\right) \quad[\omega=2 \pi f]$ $E=10 \cos \left(\frac{\pi}{6}\right)$ $E=10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3} V$
Manipal UGET-2010
Alternating Current
154982
On an $\mathrm{AC}$ circuit, the hot wire ammeter reads current $10 \mathrm{~A}$. Its peak value is
1 $10 \mathrm{~A}$
2 $20 \mathrm{~A}$
3 $14.14 \mathrm{~A}$
4 $7.07 \mathrm{~A}$
Explanation:
C The ammeter reads the rms value of current. So, $\quad \mathrm{I}_{\mathrm{rms}}=10 \mathrm{~A}$ $\\ \text { Irms } =\frac{\mathrm{I}_{\text {peak }}}{\sqrt{2}}$ $\mathrm{I}_{\text {peak }} =\sqrt{2} \mathrm{I}_{\text {rms }}$ $\mathrm{I}_{\text {peak }} =1.414 \times 10$ $\mathrm{I}_{\text {peak }} =14.14 \mathrm{~A}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
154979
A $10 \mu \mathrm{F}$ capacitor is connected across a $200 \mathrm{~V}$, $50 \mathrm{~Hz}$ AC supply. The peak current through the circuit is :
1 $0.6 \sqrt{2} \mathrm{~A}$
2 $0.6 \mathrm{~A}$
3 $\frac{0.6 \pi}{2} \mathrm{~A}$
4 $\frac{0.6}{\sqrt{2}} \mathrm{~A}$
Explanation:
A Given, Capacitor, $\mathrm{C}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}$ Voltage,$\mathrm{V}=200 \mathrm{~V}$ Frequency of AC supply $=50 \mathrm{~Hz}$ We know that, impedance of a capacitor is - $X_{C}=\frac{1}{\omega C}, \quad X_{C}=\frac{1}{2 \pi f C}$ $X_{C}=\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}$ Rms current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}$ $I=\frac{200}{\frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}}}$ $I=200 \times 2 \pi \times 50 \times 10 \times 10^{-6}$ $I=0.628 \mathrm{~A}$ Peak current, $\mathrm{I}_{\max }=\sqrt{2} \mathrm{I}=\sqrt{2} \times 0.628$ $=0.628 \sqrt{2} \mathrm{~A} \approx 0.6 \sqrt{2} \mathrm{~A}$
JCECE-2004
Alternating Current
154980
The alternating voltage and current in an electric circuit are respectively given by $E=100 \sin 100 \pi \mathrm{t}, \mathrm{I}=5 \sin 100 \pi \mathrm{t}$ The reactance of the circuit will be
1 $1 \Omega$
2 $0.05 \Omega$
3 $20 \Omega$
4 zero
Explanation:
C Given that, Alternating voltage, $\mathrm{E}=100 \sin 100 \pi \mathrm{t}$ Current, $\mathrm{I}=5 \sin 100 \pi \mathrm{t}$ From equation (i) \& (ii), we get- $\mathrm{E}_{0}=100 \mathrm{~V}$ $\mathrm{I}_{0}=5 \mathrm{~A}$ $\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~V}$ And, $\quad I_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{5}{\sqrt{2}}$ Now, we know that- $\text { Reactance of circuit }(\mathrm{R}) =\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{I}_{\mathrm{rms}}}$ $\mathrm{R} =\frac{100 / \sqrt{2}}{5 / \sqrt{2}}=20 \Omega$
Manipal UGET-2015
Alternating Current
154981
The peak value of an alternating emf $E$ given by $E=E_{0} \cos \omega t$ is $10 \mathrm{~V}$ and its frequency is 50 Hz. At a time $t=\left(\frac{1}{600}\right) s$, the instantaneous value of the emf is
1 $10 \mathrm{~V}$
2 $5 \sqrt{3} \mathrm{~V}$
3 $5 \mathrm{~V}$
4 $1 \mathrm{~V}$
Explanation:
B Given that, Voltage, $\mathrm{E}_{0}=10 \mathrm{~V}$ Frequency $(\mathrm{f})=50 \mathrm{~Hz}$ Time $(\mathrm{t})=\frac{1}{600} \mathrm{~s}$ Now, instantaneous value of emf, $E=E_{0} \cos \omega \mathrm{t}=E_{0} \cos (2 \pi f t)$ $E=10 \cos \left(2 \pi \times 50 \times \frac{1}{600}\right) \quad[\omega=2 \pi f]$ $E=10 \cos \left(\frac{\pi}{6}\right)$ $E=10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3} V$
Manipal UGET-2010
Alternating Current
154982
On an $\mathrm{AC}$ circuit, the hot wire ammeter reads current $10 \mathrm{~A}$. Its peak value is
1 $10 \mathrm{~A}$
2 $20 \mathrm{~A}$
3 $14.14 \mathrm{~A}$
4 $7.07 \mathrm{~A}$
Explanation:
C The ammeter reads the rms value of current. So, $\quad \mathrm{I}_{\mathrm{rms}}=10 \mathrm{~A}$ $\\ \text { Irms } =\frac{\mathrm{I}_{\text {peak }}}{\sqrt{2}}$ $\mathrm{I}_{\text {peak }} =\sqrt{2} \mathrm{I}_{\text {rms }}$ $\mathrm{I}_{\text {peak }} =1.414 \times 10$ $\mathrm{I}_{\text {peak }} =14.14 \mathrm{~A}$
