155199
A coil of inductance $8.4 \mathrm{mH}$ and resistance $6 \Omega$ is connected to $12 \mathrm{~V}$ battery. The current in the coil is $1 \mathrm{~A}$ at approximately the time
1 $500 \mathrm{~s}$
2 $20 \mathrm{~s}$
3 $35 \mathrm{~ms}$
4 $1 \mathrm{~ms}$
Explanation:
D Given that, Inductance $\mathrm{L}=8.4 \mathrm{mH}=8.4 \times 10^{-3} \mathrm{H}$ Resistance $\mathrm{R}=6 \Omega$, Voltage $\mathrm{V}=12 \mathrm{~V}$, Current $\mathrm{I}(\mathrm{t})=1 \mathrm{~A}$, For DC L-R circuit current through the inductor can be given by - $I(t)=I_{o}\left(1-e^{-t / \tau}\right)$ $I_{o}=\frac{V}{R}=\frac{12}{6}=2 A,$ $\tau=\frac{L}{R}=\frac{8.4 \times 10^{-3}}{6}=1.4 \times 10^{-3}$ Put the value of equation (i) $1=2\left(1-\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}\right)$ $1-\frac{1}{2}=\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}$ $\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}=\frac{1}{2}$ Taking log of both side, $\frac{-t}{1.4 \times 10^{-3}}=\log \frac{1}{2}=-0.693$ $t=0.97 \times 10^{-3}$ $t = 1 \mathrm{~m} \mathrm{sec}$
UPSEE - 2012
Alternating Current
155200
An AC source is connected in parallel with an L-C-R circuit as shown. Let $I_{S}, I_{L}, I_{C}$ and $I_{R}$ denote the currents through and $V_{S}, V_{L}, V_{C}$ and $V_{R}$ the voltages across the corresponding components. Then,
D Given circuit diagram - For the L-C-R circuit - $I_{S}=\sqrt{I_{R}^{2}+\left(I_{L}-I_{C}\right)^{2}}$ At resonance $\mathrm{I}_{\mathrm{L}}=\mathrm{I}_{\mathrm{C}}$ So, $\quad I_{S}=I_{R}$ Here $I_{L}$ and $I_{C}$ can be much greater than $I_{R}$. So $I_{L}, I_{C}$ may be greater than $\mathrm{I}_{\mathrm{S}}$.
UPSEE - 2012
Alternating Current
155201
Figure represents two bulbs $B_{1}$ and $B_{2}$, resistor $R$ and an inductor $L$. When the switch $S$ is turned off
1 both $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ die out promptly
2 both $B_{1}$ and $B_{2}$ die out with some delay
3 $B_{1}$ dies out promptly but $B_{2}$ with some delay
4 $\mathrm{B}_{2}$ dies out promptly but $\mathrm{B}_{1}$ with some delay
Explanation:
C Given circuit diagram - When the switch is turned off, current decreases but not suddenly zero, according to lenz's law emf is induced in inductor, which opposes the decrease in current. Since current does not become zero rapidly so, bulb $\mathrm{B}_{2}$ dies out with some delay. On the other hand bulb $\mathrm{B}_{1}$ and resistance $\mathrm{R}$, constitute a simple $\mathrm{R}$ circuit, in which as the voltage becomes zero, current becomes zero and $\mathrm{B}_{1}$ dies out promptly.
UPSEE - 2012
Alternating Current
155202
What is the value of inductance $L$ for which the current is a maximum in a series LCR circuit with $C=10 \mu \mathrm{F}$ and $\omega=1000$ rad. $\mathrm{s}^{-1}$ ?
1 $100 \mathrm{mH}$
2 $1 \mathrm{mH}$
3 Cannot be calculated unless $\mathrm{R}$ is known
4 $10 \mathrm{mH}$
Explanation:
A Given that, Capacitor $\mathrm{C}=10 \mu \mathrm{F},=10 \times 10^{-6} \mathrm{~F}$ Angular velocity $\omega=1000 \mathrm{rad} / \mathrm{sec}$ $\mathrm{L}=$ ? For an L-C-R series circuit current will be maximum at resonance As we know that resonance frequency $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \quad \omega^{2}=\frac{1}{\mathrm{LC}}$ $\mathrm{L}=\frac{1}{\omega^{2} \mathrm{C}}$ $\because \quad \frac{1}{1000 \times 1000 \times 10 \times 10^{-6}}$ $\mathrm{~L}=0.1 \mathrm{H}=100 \mathrm{mH} .$
155199
A coil of inductance $8.4 \mathrm{mH}$ and resistance $6 \Omega$ is connected to $12 \mathrm{~V}$ battery. The current in the coil is $1 \mathrm{~A}$ at approximately the time
1 $500 \mathrm{~s}$
2 $20 \mathrm{~s}$
3 $35 \mathrm{~ms}$
4 $1 \mathrm{~ms}$
Explanation:
D Given that, Inductance $\mathrm{L}=8.4 \mathrm{mH}=8.4 \times 10^{-3} \mathrm{H}$ Resistance $\mathrm{R}=6 \Omega$, Voltage $\mathrm{V}=12 \mathrm{~V}$, Current $\mathrm{I}(\mathrm{t})=1 \mathrm{~A}$, For DC L-R circuit current through the inductor can be given by - $I(t)=I_{o}\left(1-e^{-t / \tau}\right)$ $I_{o}=\frac{V}{R}=\frac{12}{6}=2 A,$ $\tau=\frac{L}{R}=\frac{8.4 \times 10^{-3}}{6}=1.4 \times 10^{-3}$ Put the value of equation (i) $1=2\left(1-\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}\right)$ $1-\frac{1}{2}=\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}$ $\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}=\frac{1}{2}$ Taking log of both side, $\frac{-t}{1.4 \times 10^{-3}}=\log \frac{1}{2}=-0.693$ $t=0.97 \times 10^{-3}$ $t = 1 \mathrm{~m} \mathrm{sec}$
UPSEE - 2012
Alternating Current
155200
An AC source is connected in parallel with an L-C-R circuit as shown. Let $I_{S}, I_{L}, I_{C}$ and $I_{R}$ denote the currents through and $V_{S}, V_{L}, V_{C}$ and $V_{R}$ the voltages across the corresponding components. Then,
D Given circuit diagram - For the L-C-R circuit - $I_{S}=\sqrt{I_{R}^{2}+\left(I_{L}-I_{C}\right)^{2}}$ At resonance $\mathrm{I}_{\mathrm{L}}=\mathrm{I}_{\mathrm{C}}$ So, $\quad I_{S}=I_{R}$ Here $I_{L}$ and $I_{C}$ can be much greater than $I_{R}$. So $I_{L}, I_{C}$ may be greater than $\mathrm{I}_{\mathrm{S}}$.
UPSEE - 2012
Alternating Current
155201
Figure represents two bulbs $B_{1}$ and $B_{2}$, resistor $R$ and an inductor $L$. When the switch $S$ is turned off
1 both $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ die out promptly
2 both $B_{1}$ and $B_{2}$ die out with some delay
3 $B_{1}$ dies out promptly but $B_{2}$ with some delay
4 $\mathrm{B}_{2}$ dies out promptly but $\mathrm{B}_{1}$ with some delay
Explanation:
C Given circuit diagram - When the switch is turned off, current decreases but not suddenly zero, according to lenz's law emf is induced in inductor, which opposes the decrease in current. Since current does not become zero rapidly so, bulb $\mathrm{B}_{2}$ dies out with some delay. On the other hand bulb $\mathrm{B}_{1}$ and resistance $\mathrm{R}$, constitute a simple $\mathrm{R}$ circuit, in which as the voltage becomes zero, current becomes zero and $\mathrm{B}_{1}$ dies out promptly.
UPSEE - 2012
Alternating Current
155202
What is the value of inductance $L$ for which the current is a maximum in a series LCR circuit with $C=10 \mu \mathrm{F}$ and $\omega=1000$ rad. $\mathrm{s}^{-1}$ ?
1 $100 \mathrm{mH}$
2 $1 \mathrm{mH}$
3 Cannot be calculated unless $\mathrm{R}$ is known
4 $10 \mathrm{mH}$
Explanation:
A Given that, Capacitor $\mathrm{C}=10 \mu \mathrm{F},=10 \times 10^{-6} \mathrm{~F}$ Angular velocity $\omega=1000 \mathrm{rad} / \mathrm{sec}$ $\mathrm{L}=$ ? For an L-C-R series circuit current will be maximum at resonance As we know that resonance frequency $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \quad \omega^{2}=\frac{1}{\mathrm{LC}}$ $\mathrm{L}=\frac{1}{\omega^{2} \mathrm{C}}$ $\because \quad \frac{1}{1000 \times 1000 \times 10 \times 10^{-6}}$ $\mathrm{~L}=0.1 \mathrm{H}=100 \mathrm{mH} .$
155199
A coil of inductance $8.4 \mathrm{mH}$ and resistance $6 \Omega$ is connected to $12 \mathrm{~V}$ battery. The current in the coil is $1 \mathrm{~A}$ at approximately the time
1 $500 \mathrm{~s}$
2 $20 \mathrm{~s}$
3 $35 \mathrm{~ms}$
4 $1 \mathrm{~ms}$
Explanation:
D Given that, Inductance $\mathrm{L}=8.4 \mathrm{mH}=8.4 \times 10^{-3} \mathrm{H}$ Resistance $\mathrm{R}=6 \Omega$, Voltage $\mathrm{V}=12 \mathrm{~V}$, Current $\mathrm{I}(\mathrm{t})=1 \mathrm{~A}$, For DC L-R circuit current through the inductor can be given by - $I(t)=I_{o}\left(1-e^{-t / \tau}\right)$ $I_{o}=\frac{V}{R}=\frac{12}{6}=2 A,$ $\tau=\frac{L}{R}=\frac{8.4 \times 10^{-3}}{6}=1.4 \times 10^{-3}$ Put the value of equation (i) $1=2\left(1-\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}\right)$ $1-\frac{1}{2}=\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}$ $\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}=\frac{1}{2}$ Taking log of both side, $\frac{-t}{1.4 \times 10^{-3}}=\log \frac{1}{2}=-0.693$ $t=0.97 \times 10^{-3}$ $t = 1 \mathrm{~m} \mathrm{sec}$
UPSEE - 2012
Alternating Current
155200
An AC source is connected in parallel with an L-C-R circuit as shown. Let $I_{S}, I_{L}, I_{C}$ and $I_{R}$ denote the currents through and $V_{S}, V_{L}, V_{C}$ and $V_{R}$ the voltages across the corresponding components. Then,
D Given circuit diagram - For the L-C-R circuit - $I_{S}=\sqrt{I_{R}^{2}+\left(I_{L}-I_{C}\right)^{2}}$ At resonance $\mathrm{I}_{\mathrm{L}}=\mathrm{I}_{\mathrm{C}}$ So, $\quad I_{S}=I_{R}$ Here $I_{L}$ and $I_{C}$ can be much greater than $I_{R}$. So $I_{L}, I_{C}$ may be greater than $\mathrm{I}_{\mathrm{S}}$.
UPSEE - 2012
Alternating Current
155201
Figure represents two bulbs $B_{1}$ and $B_{2}$, resistor $R$ and an inductor $L$. When the switch $S$ is turned off
1 both $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ die out promptly
2 both $B_{1}$ and $B_{2}$ die out with some delay
3 $B_{1}$ dies out promptly but $B_{2}$ with some delay
4 $\mathrm{B}_{2}$ dies out promptly but $\mathrm{B}_{1}$ with some delay
Explanation:
C Given circuit diagram - When the switch is turned off, current decreases but not suddenly zero, according to lenz's law emf is induced in inductor, which opposes the decrease in current. Since current does not become zero rapidly so, bulb $\mathrm{B}_{2}$ dies out with some delay. On the other hand bulb $\mathrm{B}_{1}$ and resistance $\mathrm{R}$, constitute a simple $\mathrm{R}$ circuit, in which as the voltage becomes zero, current becomes zero and $\mathrm{B}_{1}$ dies out promptly.
UPSEE - 2012
Alternating Current
155202
What is the value of inductance $L$ for which the current is a maximum in a series LCR circuit with $C=10 \mu \mathrm{F}$ and $\omega=1000$ rad. $\mathrm{s}^{-1}$ ?
1 $100 \mathrm{mH}$
2 $1 \mathrm{mH}$
3 Cannot be calculated unless $\mathrm{R}$ is known
4 $10 \mathrm{mH}$
Explanation:
A Given that, Capacitor $\mathrm{C}=10 \mu \mathrm{F},=10 \times 10^{-6} \mathrm{~F}$ Angular velocity $\omega=1000 \mathrm{rad} / \mathrm{sec}$ $\mathrm{L}=$ ? For an L-C-R series circuit current will be maximum at resonance As we know that resonance frequency $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \quad \omega^{2}=\frac{1}{\mathrm{LC}}$ $\mathrm{L}=\frac{1}{\omega^{2} \mathrm{C}}$ $\because \quad \frac{1}{1000 \times 1000 \times 10 \times 10^{-6}}$ $\mathrm{~L}=0.1 \mathrm{H}=100 \mathrm{mH} .$
155199
A coil of inductance $8.4 \mathrm{mH}$ and resistance $6 \Omega$ is connected to $12 \mathrm{~V}$ battery. The current in the coil is $1 \mathrm{~A}$ at approximately the time
1 $500 \mathrm{~s}$
2 $20 \mathrm{~s}$
3 $35 \mathrm{~ms}$
4 $1 \mathrm{~ms}$
Explanation:
D Given that, Inductance $\mathrm{L}=8.4 \mathrm{mH}=8.4 \times 10^{-3} \mathrm{H}$ Resistance $\mathrm{R}=6 \Omega$, Voltage $\mathrm{V}=12 \mathrm{~V}$, Current $\mathrm{I}(\mathrm{t})=1 \mathrm{~A}$, For DC L-R circuit current through the inductor can be given by - $I(t)=I_{o}\left(1-e^{-t / \tau}\right)$ $I_{o}=\frac{V}{R}=\frac{12}{6}=2 A,$ $\tau=\frac{L}{R}=\frac{8.4 \times 10^{-3}}{6}=1.4 \times 10^{-3}$ Put the value of equation (i) $1=2\left(1-\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}\right)$ $1-\frac{1}{2}=\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}$ $\mathrm{e}^{-\mathrm{t} / 1.4 \times 10^{-3}}=\frac{1}{2}$ Taking log of both side, $\frac{-t}{1.4 \times 10^{-3}}=\log \frac{1}{2}=-0.693$ $t=0.97 \times 10^{-3}$ $t = 1 \mathrm{~m} \mathrm{sec}$
UPSEE - 2012
Alternating Current
155200
An AC source is connected in parallel with an L-C-R circuit as shown. Let $I_{S}, I_{L}, I_{C}$ and $I_{R}$ denote the currents through and $V_{S}, V_{L}, V_{C}$ and $V_{R}$ the voltages across the corresponding components. Then,
D Given circuit diagram - For the L-C-R circuit - $I_{S}=\sqrt{I_{R}^{2}+\left(I_{L}-I_{C}\right)^{2}}$ At resonance $\mathrm{I}_{\mathrm{L}}=\mathrm{I}_{\mathrm{C}}$ So, $\quad I_{S}=I_{R}$ Here $I_{L}$ and $I_{C}$ can be much greater than $I_{R}$. So $I_{L}, I_{C}$ may be greater than $\mathrm{I}_{\mathrm{S}}$.
UPSEE - 2012
Alternating Current
155201
Figure represents two bulbs $B_{1}$ and $B_{2}$, resistor $R$ and an inductor $L$. When the switch $S$ is turned off
1 both $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ die out promptly
2 both $B_{1}$ and $B_{2}$ die out with some delay
3 $B_{1}$ dies out promptly but $B_{2}$ with some delay
4 $\mathrm{B}_{2}$ dies out promptly but $\mathrm{B}_{1}$ with some delay
Explanation:
C Given circuit diagram - When the switch is turned off, current decreases but not suddenly zero, according to lenz's law emf is induced in inductor, which opposes the decrease in current. Since current does not become zero rapidly so, bulb $\mathrm{B}_{2}$ dies out with some delay. On the other hand bulb $\mathrm{B}_{1}$ and resistance $\mathrm{R}$, constitute a simple $\mathrm{R}$ circuit, in which as the voltage becomes zero, current becomes zero and $\mathrm{B}_{1}$ dies out promptly.
UPSEE - 2012
Alternating Current
155202
What is the value of inductance $L$ for which the current is a maximum in a series LCR circuit with $C=10 \mu \mathrm{F}$ and $\omega=1000$ rad. $\mathrm{s}^{-1}$ ?
1 $100 \mathrm{mH}$
2 $1 \mathrm{mH}$
3 Cannot be calculated unless $\mathrm{R}$ is known
4 $10 \mathrm{mH}$
Explanation:
A Given that, Capacitor $\mathrm{C}=10 \mu \mathrm{F},=10 \times 10^{-6} \mathrm{~F}$ Angular velocity $\omega=1000 \mathrm{rad} / \mathrm{sec}$ $\mathrm{L}=$ ? For an L-C-R series circuit current will be maximum at resonance As we know that resonance frequency $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\therefore \quad \omega^{2}=\frac{1}{\mathrm{LC}}$ $\mathrm{L}=\frac{1}{\omega^{2} \mathrm{C}}$ $\because \quad \frac{1}{1000 \times 1000 \times 10 \times 10^{-6}}$ $\mathrm{~L}=0.1 \mathrm{H}=100 \mathrm{mH} .$
