NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
154967
In an LCR series circuit ( of inductance $L$, capacitance $C$ and resistance $R$ ), the impedance is minimum when the angular frequency of the source is given by
1 $\sqrt{\mathrm{LC}}$
2 $\frac{1}{\sqrt{\mathrm{LC}}}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$
5 $\sqrt{\mathrm{LCR}}$
Explanation:
B In LCR circuit, At the resonance condition, Capacitive reactance $=$ Inductive reactance $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$ $\frac{1}{\omega \mathrm{C}}=\omega \mathrm{L}$ $\omega^{2}=\frac{1}{\mathrm{LC}}$ $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ Hence, when angular frequency $(\omega)=\frac{1}{\sqrt{\mathrm{LC}}}$ then impedance is minimum.
Kerala CEE-2019
Alternating Current
154968
The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value?
1 $5 \times 10^{-3} \mathrm{~s}$
2 $2.5 \times 10^{-3} \mathrm{~s}$
3 $0.02 \mathrm{~s}$
4 $10 \times 10^{-3} \mathrm{~s}$
Explanation:
B Given, Frequency (f) $=50 \mathrm{~Hz}$ We know that, $\mathrm{i}=\mathrm{i}_{0} \cos \omega \mathrm{t}$ $\frac{\mathrm{i}_{0}}{\sqrt{2}}=\mathrm{i}_{0} \cos \omega \mathrm{t}$ $\frac{1}{\sqrt{2}}=\cos \omega \mathrm{t}$ $\frac{1}{\sqrt{2}}=\cos 2 \pi \mathrm{ft}$ $\cos \frac{\pi}{4}=\cos 2 \pi \mathrm{ft} \quad\left[\because \frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}\right]$ $\frac{\pi}{4}=2 \pi \times 50 \mathrm{t}$ $\mathrm{t}=\frac{1}{8 \times 50}=2.5 \times 10^{-3} \mathrm{~s}$ Hence, time required $\mathrm{t}=2.5 \times 10^{-3} \mathrm{~s}$ for current to reach its peak value from rms value.
Karnataka CET-2019
Alternating Current
154970
A $100 \mathrm{~W}$ bulb is connected to an $\mathrm{AC}$ source of $220 \mathrm{~V}, 50 \mathrm{~Hz}$. Then, the current flowing through the bulb is
1 $\frac{5}{11} \mathrm{~A}$
2 $\frac{1}{2} \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $\frac{3}{4} \mathrm{~A}$
Explanation:
A Given, Voltage, $\mathrm{V}=220 \mathrm{~V}$ Frequency, $\mathrm{f}=50 \mathrm{~Hz}$ Power, $\mathrm{P}=100 \mathrm{~W}$ We know that, $\mathrm{P}=\mathrm{V} \times \mathrm{I}$ $100=220 \times \mathrm{I}$ $\mathrm{I}=\frac{100}{220}=\frac{5}{11} \mathrm{~A}$ Hence, the current flowing through the bulb, $\mathrm{I}=\frac{5}{11} \mathrm{~A}$.
Karnataka CET-2018
Alternating Current
154971
The phase difference between the current and voltage is
1 $45^{\circ}$
2 $30^{\circ}$
3 $50^{\circ}$
4 $90^{\circ}$
Explanation:
A We have, AC voltage, $\mathrm{V}=2 \cos (100 \mathrm{t}+\phi)$ Inductance, $\mathrm{L}=20 \mathrm{mH}$ Resistance, $\mathrm{R}=2 \Omega$ Phase difference in LR circuit, Phase difference, $\phi=\tan ^{-1}\left(\frac{100 \times 20 \times 10^{-3}}{2}\right)$ $\phi=\tan ^{-1} 1$ $\phi=45^{\circ}$ Hence, the phase difference between voltage and current, $\phi=45^{\circ}$.
154967
In an LCR series circuit ( of inductance $L$, capacitance $C$ and resistance $R$ ), the impedance is minimum when the angular frequency of the source is given by
1 $\sqrt{\mathrm{LC}}$
2 $\frac{1}{\sqrt{\mathrm{LC}}}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$
5 $\sqrt{\mathrm{LCR}}$
Explanation:
B In LCR circuit, At the resonance condition, Capacitive reactance $=$ Inductive reactance $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$ $\frac{1}{\omega \mathrm{C}}=\omega \mathrm{L}$ $\omega^{2}=\frac{1}{\mathrm{LC}}$ $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ Hence, when angular frequency $(\omega)=\frac{1}{\sqrt{\mathrm{LC}}}$ then impedance is minimum.
Kerala CEE-2019
Alternating Current
154968
The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value?
1 $5 \times 10^{-3} \mathrm{~s}$
2 $2.5 \times 10^{-3} \mathrm{~s}$
3 $0.02 \mathrm{~s}$
4 $10 \times 10^{-3} \mathrm{~s}$
Explanation:
B Given, Frequency (f) $=50 \mathrm{~Hz}$ We know that, $\mathrm{i}=\mathrm{i}_{0} \cos \omega \mathrm{t}$ $\frac{\mathrm{i}_{0}}{\sqrt{2}}=\mathrm{i}_{0} \cos \omega \mathrm{t}$ $\frac{1}{\sqrt{2}}=\cos \omega \mathrm{t}$ $\frac{1}{\sqrt{2}}=\cos 2 \pi \mathrm{ft}$ $\cos \frac{\pi}{4}=\cos 2 \pi \mathrm{ft} \quad\left[\because \frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}\right]$ $\frac{\pi}{4}=2 \pi \times 50 \mathrm{t}$ $\mathrm{t}=\frac{1}{8 \times 50}=2.5 \times 10^{-3} \mathrm{~s}$ Hence, time required $\mathrm{t}=2.5 \times 10^{-3} \mathrm{~s}$ for current to reach its peak value from rms value.
Karnataka CET-2019
Alternating Current
154970
A $100 \mathrm{~W}$ bulb is connected to an $\mathrm{AC}$ source of $220 \mathrm{~V}, 50 \mathrm{~Hz}$. Then, the current flowing through the bulb is
1 $\frac{5}{11} \mathrm{~A}$
2 $\frac{1}{2} \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $\frac{3}{4} \mathrm{~A}$
Explanation:
A Given, Voltage, $\mathrm{V}=220 \mathrm{~V}$ Frequency, $\mathrm{f}=50 \mathrm{~Hz}$ Power, $\mathrm{P}=100 \mathrm{~W}$ We know that, $\mathrm{P}=\mathrm{V} \times \mathrm{I}$ $100=220 \times \mathrm{I}$ $\mathrm{I}=\frac{100}{220}=\frac{5}{11} \mathrm{~A}$ Hence, the current flowing through the bulb, $\mathrm{I}=\frac{5}{11} \mathrm{~A}$.
Karnataka CET-2018
Alternating Current
154971
The phase difference between the current and voltage is
1 $45^{\circ}$
2 $30^{\circ}$
3 $50^{\circ}$
4 $90^{\circ}$
Explanation:
A We have, AC voltage, $\mathrm{V}=2 \cos (100 \mathrm{t}+\phi)$ Inductance, $\mathrm{L}=20 \mathrm{mH}$ Resistance, $\mathrm{R}=2 \Omega$ Phase difference in LR circuit, Phase difference, $\phi=\tan ^{-1}\left(\frac{100 \times 20 \times 10^{-3}}{2}\right)$ $\phi=\tan ^{-1} 1$ $\phi=45^{\circ}$ Hence, the phase difference between voltage and current, $\phi=45^{\circ}$.
154967
In an LCR series circuit ( of inductance $L$, capacitance $C$ and resistance $R$ ), the impedance is minimum when the angular frequency of the source is given by
1 $\sqrt{\mathrm{LC}}$
2 $\frac{1}{\sqrt{\mathrm{LC}}}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$
5 $\sqrt{\mathrm{LCR}}$
Explanation:
B In LCR circuit, At the resonance condition, Capacitive reactance $=$ Inductive reactance $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$ $\frac{1}{\omega \mathrm{C}}=\omega \mathrm{L}$ $\omega^{2}=\frac{1}{\mathrm{LC}}$ $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ Hence, when angular frequency $(\omega)=\frac{1}{\sqrt{\mathrm{LC}}}$ then impedance is minimum.
Kerala CEE-2019
Alternating Current
154968
The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value?
1 $5 \times 10^{-3} \mathrm{~s}$
2 $2.5 \times 10^{-3} \mathrm{~s}$
3 $0.02 \mathrm{~s}$
4 $10 \times 10^{-3} \mathrm{~s}$
Explanation:
B Given, Frequency (f) $=50 \mathrm{~Hz}$ We know that, $\mathrm{i}=\mathrm{i}_{0} \cos \omega \mathrm{t}$ $\frac{\mathrm{i}_{0}}{\sqrt{2}}=\mathrm{i}_{0} \cos \omega \mathrm{t}$ $\frac{1}{\sqrt{2}}=\cos \omega \mathrm{t}$ $\frac{1}{\sqrt{2}}=\cos 2 \pi \mathrm{ft}$ $\cos \frac{\pi}{4}=\cos 2 \pi \mathrm{ft} \quad\left[\because \frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}\right]$ $\frac{\pi}{4}=2 \pi \times 50 \mathrm{t}$ $\mathrm{t}=\frac{1}{8 \times 50}=2.5 \times 10^{-3} \mathrm{~s}$ Hence, time required $\mathrm{t}=2.5 \times 10^{-3} \mathrm{~s}$ for current to reach its peak value from rms value.
Karnataka CET-2019
Alternating Current
154970
A $100 \mathrm{~W}$ bulb is connected to an $\mathrm{AC}$ source of $220 \mathrm{~V}, 50 \mathrm{~Hz}$. Then, the current flowing through the bulb is
1 $\frac{5}{11} \mathrm{~A}$
2 $\frac{1}{2} \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $\frac{3}{4} \mathrm{~A}$
Explanation:
A Given, Voltage, $\mathrm{V}=220 \mathrm{~V}$ Frequency, $\mathrm{f}=50 \mathrm{~Hz}$ Power, $\mathrm{P}=100 \mathrm{~W}$ We know that, $\mathrm{P}=\mathrm{V} \times \mathrm{I}$ $100=220 \times \mathrm{I}$ $\mathrm{I}=\frac{100}{220}=\frac{5}{11} \mathrm{~A}$ Hence, the current flowing through the bulb, $\mathrm{I}=\frac{5}{11} \mathrm{~A}$.
Karnataka CET-2018
Alternating Current
154971
The phase difference between the current and voltage is
1 $45^{\circ}$
2 $30^{\circ}$
3 $50^{\circ}$
4 $90^{\circ}$
Explanation:
A We have, AC voltage, $\mathrm{V}=2 \cos (100 \mathrm{t}+\phi)$ Inductance, $\mathrm{L}=20 \mathrm{mH}$ Resistance, $\mathrm{R}=2 \Omega$ Phase difference in LR circuit, Phase difference, $\phi=\tan ^{-1}\left(\frac{100 \times 20 \times 10^{-3}}{2}\right)$ $\phi=\tan ^{-1} 1$ $\phi=45^{\circ}$ Hence, the phase difference between voltage and current, $\phi=45^{\circ}$.
154967
In an LCR series circuit ( of inductance $L$, capacitance $C$ and resistance $R$ ), the impedance is minimum when the angular frequency of the source is given by
1 $\sqrt{\mathrm{LC}}$
2 $\frac{1}{\sqrt{\mathrm{LC}}}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$
5 $\sqrt{\mathrm{LCR}}$
Explanation:
B In LCR circuit, At the resonance condition, Capacitive reactance $=$ Inductive reactance $\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$ $\frac{1}{\omega \mathrm{C}}=\omega \mathrm{L}$ $\omega^{2}=\frac{1}{\mathrm{LC}}$ $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ Hence, when angular frequency $(\omega)=\frac{1}{\sqrt{\mathrm{LC}}}$ then impedance is minimum.
Kerala CEE-2019
Alternating Current
154968
The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value?
1 $5 \times 10^{-3} \mathrm{~s}$
2 $2.5 \times 10^{-3} \mathrm{~s}$
3 $0.02 \mathrm{~s}$
4 $10 \times 10^{-3} \mathrm{~s}$
Explanation:
B Given, Frequency (f) $=50 \mathrm{~Hz}$ We know that, $\mathrm{i}=\mathrm{i}_{0} \cos \omega \mathrm{t}$ $\frac{\mathrm{i}_{0}}{\sqrt{2}}=\mathrm{i}_{0} \cos \omega \mathrm{t}$ $\frac{1}{\sqrt{2}}=\cos \omega \mathrm{t}$ $\frac{1}{\sqrt{2}}=\cos 2 \pi \mathrm{ft}$ $\cos \frac{\pi}{4}=\cos 2 \pi \mathrm{ft} \quad\left[\because \frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}\right]$ $\frac{\pi}{4}=2 \pi \times 50 \mathrm{t}$ $\mathrm{t}=\frac{1}{8 \times 50}=2.5 \times 10^{-3} \mathrm{~s}$ Hence, time required $\mathrm{t}=2.5 \times 10^{-3} \mathrm{~s}$ for current to reach its peak value from rms value.
Karnataka CET-2019
Alternating Current
154970
A $100 \mathrm{~W}$ bulb is connected to an $\mathrm{AC}$ source of $220 \mathrm{~V}, 50 \mathrm{~Hz}$. Then, the current flowing through the bulb is
1 $\frac{5}{11} \mathrm{~A}$
2 $\frac{1}{2} \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $\frac{3}{4} \mathrm{~A}$
Explanation:
A Given, Voltage, $\mathrm{V}=220 \mathrm{~V}$ Frequency, $\mathrm{f}=50 \mathrm{~Hz}$ Power, $\mathrm{P}=100 \mathrm{~W}$ We know that, $\mathrm{P}=\mathrm{V} \times \mathrm{I}$ $100=220 \times \mathrm{I}$ $\mathrm{I}=\frac{100}{220}=\frac{5}{11} \mathrm{~A}$ Hence, the current flowing through the bulb, $\mathrm{I}=\frac{5}{11} \mathrm{~A}$.
Karnataka CET-2018
Alternating Current
154971
The phase difference between the current and voltage is
1 $45^{\circ}$
2 $30^{\circ}$
3 $50^{\circ}$
4 $90^{\circ}$
Explanation:
A We have, AC voltage, $\mathrm{V}=2 \cos (100 \mathrm{t}+\phi)$ Inductance, $\mathrm{L}=20 \mathrm{mH}$ Resistance, $\mathrm{R}=2 \Omega$ Phase difference in LR circuit, Phase difference, $\phi=\tan ^{-1}\left(\frac{100 \times 20 \times 10^{-3}}{2}\right)$ $\phi=\tan ^{-1} 1$ $\phi=45^{\circ}$ Hence, the phase difference between voltage and current, $\phi=45^{\circ}$.
