NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155163
In an series LCR circuit the phase difference between voltage across $R$ and $C$ is
1 0
2 $\frac{\pi}{2}$
3 $\pi$
4 $\frac{3 \pi}{2}$ [SRM JEE-2018]
Explanation:
B In an series LCR circuit the phase difference between voltage across $\mathrm{R}$ and $\mathrm{C}$ is- As shown in figure phase difference between voltage across $\mathrm{R}$ and $\mathrm{C}$ is $\frac{\pi}{2}$
Alternating Current
155168
An alternating current is flowing through a series LCR circuit. It is found that the current reaches a value of $1 \mathrm{~mA}$ at both $200 \mathrm{~Hz}$ and 800 $\mathrm{Hz}$ frequency. What is the resonance frequency of the circuit?
1 $600 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $500 \mathrm{~Hz}$
4 $400 \mathrm{~Hz}$
Explanation:
D Given that, $\mathrm{f}_{1}=200 \mathrm{~Hz}, \mathrm{f}_{2}=800 \mathrm{~Hz}$, for current $(\mathrm{I})=1 \mathrm{~mA}$ So, resonance frequency $\left(\mathrm{f}_{0}\right)=\sqrt{\mathrm{f}_{1} \mathrm{f}_{2}}$ $\mathrm{f}_{0} =\sqrt{800 \times 200}$ $\mathrm{f}_{0} =400 \mathrm{~Hz}$
WB JEE 2018
Alternating Current
155169
A coil of 40 henry inductance is connected in series with a resistance of $8 \mathrm{ohm}$ and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is
1 20 seconds
2 5 seconds
3 $1 / 5$ seconds
4 40 seconds
Explanation:
B Given that, $\mathrm{L}=40 \mathrm{H} \text { and } \mathrm{R}=8 \Omega$ Then, time constant for LR circuit $(\tau)=\frac{L}{R}$ $\tau=\frac{40}{8}=5 \text { second }$
VITEEE-2018
Alternating Current
155170
In an oscillation of $\mathrm{L}-\mathrm{C}$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
1 $\frac{\mathrm{Q}}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{2}}$
3 $\frac{\mathrm{Q}}{\sqrt{3}}$
4 $\frac{\mathrm{Q}}{3}$
Explanation:
B If q is the required charge, Then, $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \times \frac{\mathrm{Q}^{2}}{2 \mathrm{C}}$ $\therefore \quad \mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}$ Hence, the required charge $(q)=\frac{\mathrm{Q}}{\sqrt{2}}$
155163
In an series LCR circuit the phase difference between voltage across $R$ and $C$ is
1 0
2 $\frac{\pi}{2}$
3 $\pi$
4 $\frac{3 \pi}{2}$ [SRM JEE-2018]
Explanation:
B In an series LCR circuit the phase difference between voltage across $\mathrm{R}$ and $\mathrm{C}$ is- As shown in figure phase difference between voltage across $\mathrm{R}$ and $\mathrm{C}$ is $\frac{\pi}{2}$
Alternating Current
155168
An alternating current is flowing through a series LCR circuit. It is found that the current reaches a value of $1 \mathrm{~mA}$ at both $200 \mathrm{~Hz}$ and 800 $\mathrm{Hz}$ frequency. What is the resonance frequency of the circuit?
1 $600 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $500 \mathrm{~Hz}$
4 $400 \mathrm{~Hz}$
Explanation:
D Given that, $\mathrm{f}_{1}=200 \mathrm{~Hz}, \mathrm{f}_{2}=800 \mathrm{~Hz}$, for current $(\mathrm{I})=1 \mathrm{~mA}$ So, resonance frequency $\left(\mathrm{f}_{0}\right)=\sqrt{\mathrm{f}_{1} \mathrm{f}_{2}}$ $\mathrm{f}_{0} =\sqrt{800 \times 200}$ $\mathrm{f}_{0} =400 \mathrm{~Hz}$
WB JEE 2018
Alternating Current
155169
A coil of 40 henry inductance is connected in series with a resistance of $8 \mathrm{ohm}$ and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is
1 20 seconds
2 5 seconds
3 $1 / 5$ seconds
4 40 seconds
Explanation:
B Given that, $\mathrm{L}=40 \mathrm{H} \text { and } \mathrm{R}=8 \Omega$ Then, time constant for LR circuit $(\tau)=\frac{L}{R}$ $\tau=\frac{40}{8}=5 \text { second }$
VITEEE-2018
Alternating Current
155170
In an oscillation of $\mathrm{L}-\mathrm{C}$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
1 $\frac{\mathrm{Q}}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{2}}$
3 $\frac{\mathrm{Q}}{\sqrt{3}}$
4 $\frac{\mathrm{Q}}{3}$
Explanation:
B If q is the required charge, Then, $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \times \frac{\mathrm{Q}^{2}}{2 \mathrm{C}}$ $\therefore \quad \mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}$ Hence, the required charge $(q)=\frac{\mathrm{Q}}{\sqrt{2}}$
155163
In an series LCR circuit the phase difference between voltage across $R$ and $C$ is
1 0
2 $\frac{\pi}{2}$
3 $\pi$
4 $\frac{3 \pi}{2}$ [SRM JEE-2018]
Explanation:
B In an series LCR circuit the phase difference between voltage across $\mathrm{R}$ and $\mathrm{C}$ is- As shown in figure phase difference between voltage across $\mathrm{R}$ and $\mathrm{C}$ is $\frac{\pi}{2}$
Alternating Current
155168
An alternating current is flowing through a series LCR circuit. It is found that the current reaches a value of $1 \mathrm{~mA}$ at both $200 \mathrm{~Hz}$ and 800 $\mathrm{Hz}$ frequency. What is the resonance frequency of the circuit?
1 $600 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $500 \mathrm{~Hz}$
4 $400 \mathrm{~Hz}$
Explanation:
D Given that, $\mathrm{f}_{1}=200 \mathrm{~Hz}, \mathrm{f}_{2}=800 \mathrm{~Hz}$, for current $(\mathrm{I})=1 \mathrm{~mA}$ So, resonance frequency $\left(\mathrm{f}_{0}\right)=\sqrt{\mathrm{f}_{1} \mathrm{f}_{2}}$ $\mathrm{f}_{0} =\sqrt{800 \times 200}$ $\mathrm{f}_{0} =400 \mathrm{~Hz}$
WB JEE 2018
Alternating Current
155169
A coil of 40 henry inductance is connected in series with a resistance of $8 \mathrm{ohm}$ and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is
1 20 seconds
2 5 seconds
3 $1 / 5$ seconds
4 40 seconds
Explanation:
B Given that, $\mathrm{L}=40 \mathrm{H} \text { and } \mathrm{R}=8 \Omega$ Then, time constant for LR circuit $(\tau)=\frac{L}{R}$ $\tau=\frac{40}{8}=5 \text { second }$
VITEEE-2018
Alternating Current
155170
In an oscillation of $\mathrm{L}-\mathrm{C}$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
1 $\frac{\mathrm{Q}}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{2}}$
3 $\frac{\mathrm{Q}}{\sqrt{3}}$
4 $\frac{\mathrm{Q}}{3}$
Explanation:
B If q is the required charge, Then, $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \times \frac{\mathrm{Q}^{2}}{2 \mathrm{C}}$ $\therefore \quad \mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}$ Hence, the required charge $(q)=\frac{\mathrm{Q}}{\sqrt{2}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155163
In an series LCR circuit the phase difference between voltage across $R$ and $C$ is
1 0
2 $\frac{\pi}{2}$
3 $\pi$
4 $\frac{3 \pi}{2}$ [SRM JEE-2018]
Explanation:
B In an series LCR circuit the phase difference between voltage across $\mathrm{R}$ and $\mathrm{C}$ is- As shown in figure phase difference between voltage across $\mathrm{R}$ and $\mathrm{C}$ is $\frac{\pi}{2}$
Alternating Current
155168
An alternating current is flowing through a series LCR circuit. It is found that the current reaches a value of $1 \mathrm{~mA}$ at both $200 \mathrm{~Hz}$ and 800 $\mathrm{Hz}$ frequency. What is the resonance frequency of the circuit?
1 $600 \mathrm{~Hz}$
2 $300 \mathrm{~Hz}$
3 $500 \mathrm{~Hz}$
4 $400 \mathrm{~Hz}$
Explanation:
D Given that, $\mathrm{f}_{1}=200 \mathrm{~Hz}, \mathrm{f}_{2}=800 \mathrm{~Hz}$, for current $(\mathrm{I})=1 \mathrm{~mA}$ So, resonance frequency $\left(\mathrm{f}_{0}\right)=\sqrt{\mathrm{f}_{1} \mathrm{f}_{2}}$ $\mathrm{f}_{0} =\sqrt{800 \times 200}$ $\mathrm{f}_{0} =400 \mathrm{~Hz}$
WB JEE 2018
Alternating Current
155169
A coil of 40 henry inductance is connected in series with a resistance of $8 \mathrm{ohm}$ and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is
1 20 seconds
2 5 seconds
3 $1 / 5$ seconds
4 40 seconds
Explanation:
B Given that, $\mathrm{L}=40 \mathrm{H} \text { and } \mathrm{R}=8 \Omega$ Then, time constant for LR circuit $(\tau)=\frac{L}{R}$ $\tau=\frac{40}{8}=5 \text { second }$
VITEEE-2018
Alternating Current
155170
In an oscillation of $\mathrm{L}-\mathrm{C}$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
1 $\frac{\mathrm{Q}}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{2}}$
3 $\frac{\mathrm{Q}}{\sqrt{3}}$
4 $\frac{\mathrm{Q}}{3}$
Explanation:
B If q is the required charge, Then, $\frac{\mathrm{q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \times \frac{\mathrm{Q}^{2}}{2 \mathrm{C}}$ $\therefore \quad \mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}$ Hence, the required charge $(q)=\frac{\mathrm{Q}}{\sqrt{2}}$
