155203
An inductance $L$ and a resistance $R$ are connected in series with a battery of emf $\varepsilon$. The maximum rate at which the energy is stored in the magnetic field is :
1 $\frac{\varepsilon^{2}}{4 \mathrm{R}}$
2 $\frac{\varepsilon^{2}}{2 R}$
3 $\frac{2 \mathrm{R}}{\varepsilon}$
4 $\frac{4 \mathrm{R}}{\varepsilon}$
Explanation:
A As we know that energy stored in magnetic field is given by - $\mathrm{E}=\frac{1}{2} \mathrm{Li}^{2}$ For an R-L series circuit current is - $\mathrm{i}=\mathrm{I}_{\mathrm{o}}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)$ Put the value of $I$ in equation (i), $\mathrm{E}=\frac{1}{2}\left[\operatorname{LI}_{\mathrm{o}}^{2}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)^{2}\right]$ Power $=$ Rate of energy $=\frac{\mathrm{dE}}{\mathrm{dt}}$ $P=\frac{d}{d t}\left[\frac{1}{2} \operatorname{LI}_{0}^{2}\left(1-\mathrm{e}^{-t / \tau}\right)^{2}\right]$ $=\frac{1}{2} \mathrm{LI}_{\mathrm{o}}^{2} \cdot 2\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right) \cdot\left[-\mathrm{e}^{-\mathrm{t} / \tau} \times \frac{-1}{\tau}\right]$ $\mathrm{P}=\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left(\mathrm{e}^{-\mathrm{t} / \tau}-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)$ Maximum power, $\frac{\mathrm{dP}}{\mathrm{dt}}=0$ $\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{e}^{-\mathrm{t} / \tau}-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)=0$ $\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left[\left(-\frac{1}{\tau}\right) \mathrm{e}^{-\mathrm{t} / \tau}+\frac{2}{\tau} \mathrm{e}^{-2 \mathrm{t} / \tau}\right]=0$ $\frac{1}{\tau} \mathrm{e}^{-\mathrm{t} / \tau}=\frac{2}{\tau} \mathrm{e}^{-2 \mathrm{t} / \tau}$ $\frac{\mathrm{e}^{-2 \mathrm{t} / \tau}}{\mathrm{e}^{-\mathrm{t} / \tau}}=\frac{1}{2}$ $\mathrm{e}^{-\mathrm{t} / \tau}=\frac{1}{2}$ Put the value of $\mathrm{e}^{-\mathrm{t} / \tau}=\frac{1}{2}$ in equation (ii), $\mathrm{P} =\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left[\frac{1}{2}\left(1-\frac{1}{2}\right)\right]$ $\mathrm{P} =\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau} \times \frac{1}{4}$ $=\mathrm{L}\left(\frac{\varepsilon}{\mathrm{R}}\right)^{2} \frac{1}{\mathrm{~L} / \mathrm{R}} \times \frac{1}{4} \quad\left\{\because \mathrm{I}_{\mathrm{o}}=\frac{\varepsilon}{\mathrm{R}}, \tau=\frac{\mathrm{L}}{\mathrm{R}}\right\}$ $\mathrm{P} =\frac{\varepsilon^{2}}{4 \mathrm{R}}$
UPSEE - 2006
Alternating Current
155204
A current of $i=2 \sin (\pi t / 3) \quad A$ is flowing in an inductor of $2 \mathrm{H}$. The amount of work done in increasing the current from $1.0 \mathrm{~A}$ to $2.0 \mathrm{~A}$ is :
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $3 \mathrm{~J}$
4 $4 \mathrm{~J}$
Explanation:
C Given that, Primary current $i_{1}=1 \mathrm{~A}$ Secondary current $\mathrm{i}_{2}=2 \mathrm{~A}$ Self inductance $\mathrm{L}=2 \mathrm{H}$ $\text { Work done }=\text { Change in energy } =\frac{1}{2}\left(\mathrm{Li}_{2}^{2}-\mathrm{Li}_{1}^{2}\right)$ $=\frac{1}{2} \mathrm{~L}\left(\mathrm{i}_{2}^{2}-\mathrm{i}_{1}^{2}\right)$ $=\frac{1}{2} \times 2\left[(2)^{2}-(1)^{2}\right]$ $=4-1$ Work done $=3 \mathrm{~J}$
UPSEE - 2004
Alternating Current
155206
An L-C-R series circuit is at resonance. Then
1 The phase difference between current and voltage is $90^{\circ}$
2 The phase difference between current and voltage is $45^{\circ}$
3 Its impedance is purely resistive
4 Its impedance is zero
5 The current is minimum
Explanation:
C As we know that, at resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ So, impedance $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+0}$ $\mathrm{Z}=\mathrm{R}$ Hence, In an LCR series circuit at resonance impedance is purely resistive.
Kerala CEE - 2010
Alternating Current
155207
An L-C-R series AC circuit is at resonance with $10 \mathrm{~V}$ each across $\mathrm{L}, \mathrm{C}$ and $\mathrm{R}$. If the resistance is halved, the respective voltages across $L, C$ and $R$ are
1 $10 \mathrm{~V}, 10 \mathrm{~V}$ and $5 \mathrm{~V}$
2 $10 \mathrm{~V}, 10 \mathrm{~V}$ and $10 \mathrm{~V}$
3 $20 \mathrm{~V}, 20 \mathrm{~V}$ and $5 \mathrm{~V}$
4 $20 \mathrm{~V}, 20 \mathrm{~V}$ and $10 \mathrm{~V}$
5 $5 \mathrm{~V}, 5 \mathrm{~V}$ and $5 \mathrm{~V}$
Explanation:
D Given, Voltage (v) $=10$ volt Let, Initial current $=\mathrm{i}$ Initial resistance $=\mathrm{R}$ Now, Inductive resistance $=\mathrm{X}_{\mathrm{L}}$ Capacitive resistance $=\mathrm{X}_{\mathrm{C}}$ We know that, When resistance is halved current will be doubled $\therefore \quad \mathrm{v}_{\mathrm{R}}=2 \mathrm{i} \times(\mathrm{R} / 2)=\mathrm{iR}=10 \mathrm{~V}$ $\mathrm{v}_{\mathrm{C}}=2 \mathrm{i} \times \mathrm{X}_{\mathrm{C}}=20 \mathrm{~V}$ $\mathrm{v}_{\mathrm{L}}=2 \mathrm{i} \times \mathrm{X}_{\mathrm{L}}=20 \mathrm{~V}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155203
An inductance $L$ and a resistance $R$ are connected in series with a battery of emf $\varepsilon$. The maximum rate at which the energy is stored in the magnetic field is :
1 $\frac{\varepsilon^{2}}{4 \mathrm{R}}$
2 $\frac{\varepsilon^{2}}{2 R}$
3 $\frac{2 \mathrm{R}}{\varepsilon}$
4 $\frac{4 \mathrm{R}}{\varepsilon}$
Explanation:
A As we know that energy stored in magnetic field is given by - $\mathrm{E}=\frac{1}{2} \mathrm{Li}^{2}$ For an R-L series circuit current is - $\mathrm{i}=\mathrm{I}_{\mathrm{o}}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)$ Put the value of $I$ in equation (i), $\mathrm{E}=\frac{1}{2}\left[\operatorname{LI}_{\mathrm{o}}^{2}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)^{2}\right]$ Power $=$ Rate of energy $=\frac{\mathrm{dE}}{\mathrm{dt}}$ $P=\frac{d}{d t}\left[\frac{1}{2} \operatorname{LI}_{0}^{2}\left(1-\mathrm{e}^{-t / \tau}\right)^{2}\right]$ $=\frac{1}{2} \mathrm{LI}_{\mathrm{o}}^{2} \cdot 2\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right) \cdot\left[-\mathrm{e}^{-\mathrm{t} / \tau} \times \frac{-1}{\tau}\right]$ $\mathrm{P}=\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left(\mathrm{e}^{-\mathrm{t} / \tau}-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)$ Maximum power, $\frac{\mathrm{dP}}{\mathrm{dt}}=0$ $\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{e}^{-\mathrm{t} / \tau}-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)=0$ $\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left[\left(-\frac{1}{\tau}\right) \mathrm{e}^{-\mathrm{t} / \tau}+\frac{2}{\tau} \mathrm{e}^{-2 \mathrm{t} / \tau}\right]=0$ $\frac{1}{\tau} \mathrm{e}^{-\mathrm{t} / \tau}=\frac{2}{\tau} \mathrm{e}^{-2 \mathrm{t} / \tau}$ $\frac{\mathrm{e}^{-2 \mathrm{t} / \tau}}{\mathrm{e}^{-\mathrm{t} / \tau}}=\frac{1}{2}$ $\mathrm{e}^{-\mathrm{t} / \tau}=\frac{1}{2}$ Put the value of $\mathrm{e}^{-\mathrm{t} / \tau}=\frac{1}{2}$ in equation (ii), $\mathrm{P} =\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left[\frac{1}{2}\left(1-\frac{1}{2}\right)\right]$ $\mathrm{P} =\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau} \times \frac{1}{4}$ $=\mathrm{L}\left(\frac{\varepsilon}{\mathrm{R}}\right)^{2} \frac{1}{\mathrm{~L} / \mathrm{R}} \times \frac{1}{4} \quad\left\{\because \mathrm{I}_{\mathrm{o}}=\frac{\varepsilon}{\mathrm{R}}, \tau=\frac{\mathrm{L}}{\mathrm{R}}\right\}$ $\mathrm{P} =\frac{\varepsilon^{2}}{4 \mathrm{R}}$
UPSEE - 2006
Alternating Current
155204
A current of $i=2 \sin (\pi t / 3) \quad A$ is flowing in an inductor of $2 \mathrm{H}$. The amount of work done in increasing the current from $1.0 \mathrm{~A}$ to $2.0 \mathrm{~A}$ is :
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $3 \mathrm{~J}$
4 $4 \mathrm{~J}$
Explanation:
C Given that, Primary current $i_{1}=1 \mathrm{~A}$ Secondary current $\mathrm{i}_{2}=2 \mathrm{~A}$ Self inductance $\mathrm{L}=2 \mathrm{H}$ $\text { Work done }=\text { Change in energy } =\frac{1}{2}\left(\mathrm{Li}_{2}^{2}-\mathrm{Li}_{1}^{2}\right)$ $=\frac{1}{2} \mathrm{~L}\left(\mathrm{i}_{2}^{2}-\mathrm{i}_{1}^{2}\right)$ $=\frac{1}{2} \times 2\left[(2)^{2}-(1)^{2}\right]$ $=4-1$ Work done $=3 \mathrm{~J}$
UPSEE - 2004
Alternating Current
155206
An L-C-R series circuit is at resonance. Then
1 The phase difference between current and voltage is $90^{\circ}$
2 The phase difference between current and voltage is $45^{\circ}$
3 Its impedance is purely resistive
4 Its impedance is zero
5 The current is minimum
Explanation:
C As we know that, at resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ So, impedance $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+0}$ $\mathrm{Z}=\mathrm{R}$ Hence, In an LCR series circuit at resonance impedance is purely resistive.
Kerala CEE - 2010
Alternating Current
155207
An L-C-R series AC circuit is at resonance with $10 \mathrm{~V}$ each across $\mathrm{L}, \mathrm{C}$ and $\mathrm{R}$. If the resistance is halved, the respective voltages across $L, C$ and $R$ are
1 $10 \mathrm{~V}, 10 \mathrm{~V}$ and $5 \mathrm{~V}$
2 $10 \mathrm{~V}, 10 \mathrm{~V}$ and $10 \mathrm{~V}$
3 $20 \mathrm{~V}, 20 \mathrm{~V}$ and $5 \mathrm{~V}$
4 $20 \mathrm{~V}, 20 \mathrm{~V}$ and $10 \mathrm{~V}$
5 $5 \mathrm{~V}, 5 \mathrm{~V}$ and $5 \mathrm{~V}$
Explanation:
D Given, Voltage (v) $=10$ volt Let, Initial current $=\mathrm{i}$ Initial resistance $=\mathrm{R}$ Now, Inductive resistance $=\mathrm{X}_{\mathrm{L}}$ Capacitive resistance $=\mathrm{X}_{\mathrm{C}}$ We know that, When resistance is halved current will be doubled $\therefore \quad \mathrm{v}_{\mathrm{R}}=2 \mathrm{i} \times(\mathrm{R} / 2)=\mathrm{iR}=10 \mathrm{~V}$ $\mathrm{v}_{\mathrm{C}}=2 \mathrm{i} \times \mathrm{X}_{\mathrm{C}}=20 \mathrm{~V}$ $\mathrm{v}_{\mathrm{L}}=2 \mathrm{i} \times \mathrm{X}_{\mathrm{L}}=20 \mathrm{~V}$
155203
An inductance $L$ and a resistance $R$ are connected in series with a battery of emf $\varepsilon$. The maximum rate at which the energy is stored in the magnetic field is :
1 $\frac{\varepsilon^{2}}{4 \mathrm{R}}$
2 $\frac{\varepsilon^{2}}{2 R}$
3 $\frac{2 \mathrm{R}}{\varepsilon}$
4 $\frac{4 \mathrm{R}}{\varepsilon}$
Explanation:
A As we know that energy stored in magnetic field is given by - $\mathrm{E}=\frac{1}{2} \mathrm{Li}^{2}$ For an R-L series circuit current is - $\mathrm{i}=\mathrm{I}_{\mathrm{o}}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)$ Put the value of $I$ in equation (i), $\mathrm{E}=\frac{1}{2}\left[\operatorname{LI}_{\mathrm{o}}^{2}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)^{2}\right]$ Power $=$ Rate of energy $=\frac{\mathrm{dE}}{\mathrm{dt}}$ $P=\frac{d}{d t}\left[\frac{1}{2} \operatorname{LI}_{0}^{2}\left(1-\mathrm{e}^{-t / \tau}\right)^{2}\right]$ $=\frac{1}{2} \mathrm{LI}_{\mathrm{o}}^{2} \cdot 2\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right) \cdot\left[-\mathrm{e}^{-\mathrm{t} / \tau} \times \frac{-1}{\tau}\right]$ $\mathrm{P}=\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left(\mathrm{e}^{-\mathrm{t} / \tau}-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)$ Maximum power, $\frac{\mathrm{dP}}{\mathrm{dt}}=0$ $\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{e}^{-\mathrm{t} / \tau}-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)=0$ $\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left[\left(-\frac{1}{\tau}\right) \mathrm{e}^{-\mathrm{t} / \tau}+\frac{2}{\tau} \mathrm{e}^{-2 \mathrm{t} / \tau}\right]=0$ $\frac{1}{\tau} \mathrm{e}^{-\mathrm{t} / \tau}=\frac{2}{\tau} \mathrm{e}^{-2 \mathrm{t} / \tau}$ $\frac{\mathrm{e}^{-2 \mathrm{t} / \tau}}{\mathrm{e}^{-\mathrm{t} / \tau}}=\frac{1}{2}$ $\mathrm{e}^{-\mathrm{t} / \tau}=\frac{1}{2}$ Put the value of $\mathrm{e}^{-\mathrm{t} / \tau}=\frac{1}{2}$ in equation (ii), $\mathrm{P} =\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left[\frac{1}{2}\left(1-\frac{1}{2}\right)\right]$ $\mathrm{P} =\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau} \times \frac{1}{4}$ $=\mathrm{L}\left(\frac{\varepsilon}{\mathrm{R}}\right)^{2} \frac{1}{\mathrm{~L} / \mathrm{R}} \times \frac{1}{4} \quad\left\{\because \mathrm{I}_{\mathrm{o}}=\frac{\varepsilon}{\mathrm{R}}, \tau=\frac{\mathrm{L}}{\mathrm{R}}\right\}$ $\mathrm{P} =\frac{\varepsilon^{2}}{4 \mathrm{R}}$
UPSEE - 2006
Alternating Current
155204
A current of $i=2 \sin (\pi t / 3) \quad A$ is flowing in an inductor of $2 \mathrm{H}$. The amount of work done in increasing the current from $1.0 \mathrm{~A}$ to $2.0 \mathrm{~A}$ is :
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $3 \mathrm{~J}$
4 $4 \mathrm{~J}$
Explanation:
C Given that, Primary current $i_{1}=1 \mathrm{~A}$ Secondary current $\mathrm{i}_{2}=2 \mathrm{~A}$ Self inductance $\mathrm{L}=2 \mathrm{H}$ $\text { Work done }=\text { Change in energy } =\frac{1}{2}\left(\mathrm{Li}_{2}^{2}-\mathrm{Li}_{1}^{2}\right)$ $=\frac{1}{2} \mathrm{~L}\left(\mathrm{i}_{2}^{2}-\mathrm{i}_{1}^{2}\right)$ $=\frac{1}{2} \times 2\left[(2)^{2}-(1)^{2}\right]$ $=4-1$ Work done $=3 \mathrm{~J}$
UPSEE - 2004
Alternating Current
155206
An L-C-R series circuit is at resonance. Then
1 The phase difference between current and voltage is $90^{\circ}$
2 The phase difference between current and voltage is $45^{\circ}$
3 Its impedance is purely resistive
4 Its impedance is zero
5 The current is minimum
Explanation:
C As we know that, at resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ So, impedance $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+0}$ $\mathrm{Z}=\mathrm{R}$ Hence, In an LCR series circuit at resonance impedance is purely resistive.
Kerala CEE - 2010
Alternating Current
155207
An L-C-R series AC circuit is at resonance with $10 \mathrm{~V}$ each across $\mathrm{L}, \mathrm{C}$ and $\mathrm{R}$. If the resistance is halved, the respective voltages across $L, C$ and $R$ are
1 $10 \mathrm{~V}, 10 \mathrm{~V}$ and $5 \mathrm{~V}$
2 $10 \mathrm{~V}, 10 \mathrm{~V}$ and $10 \mathrm{~V}$
3 $20 \mathrm{~V}, 20 \mathrm{~V}$ and $5 \mathrm{~V}$
4 $20 \mathrm{~V}, 20 \mathrm{~V}$ and $10 \mathrm{~V}$
5 $5 \mathrm{~V}, 5 \mathrm{~V}$ and $5 \mathrm{~V}$
Explanation:
D Given, Voltage (v) $=10$ volt Let, Initial current $=\mathrm{i}$ Initial resistance $=\mathrm{R}$ Now, Inductive resistance $=\mathrm{X}_{\mathrm{L}}$ Capacitive resistance $=\mathrm{X}_{\mathrm{C}}$ We know that, When resistance is halved current will be doubled $\therefore \quad \mathrm{v}_{\mathrm{R}}=2 \mathrm{i} \times(\mathrm{R} / 2)=\mathrm{iR}=10 \mathrm{~V}$ $\mathrm{v}_{\mathrm{C}}=2 \mathrm{i} \times \mathrm{X}_{\mathrm{C}}=20 \mathrm{~V}$ $\mathrm{v}_{\mathrm{L}}=2 \mathrm{i} \times \mathrm{X}_{\mathrm{L}}=20 \mathrm{~V}$
155203
An inductance $L$ and a resistance $R$ are connected in series with a battery of emf $\varepsilon$. The maximum rate at which the energy is stored in the magnetic field is :
1 $\frac{\varepsilon^{2}}{4 \mathrm{R}}$
2 $\frac{\varepsilon^{2}}{2 R}$
3 $\frac{2 \mathrm{R}}{\varepsilon}$
4 $\frac{4 \mathrm{R}}{\varepsilon}$
Explanation:
A As we know that energy stored in magnetic field is given by - $\mathrm{E}=\frac{1}{2} \mathrm{Li}^{2}$ For an R-L series circuit current is - $\mathrm{i}=\mathrm{I}_{\mathrm{o}}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)$ Put the value of $I$ in equation (i), $\mathrm{E}=\frac{1}{2}\left[\operatorname{LI}_{\mathrm{o}}^{2}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)^{2}\right]$ Power $=$ Rate of energy $=\frac{\mathrm{dE}}{\mathrm{dt}}$ $P=\frac{d}{d t}\left[\frac{1}{2} \operatorname{LI}_{0}^{2}\left(1-\mathrm{e}^{-t / \tau}\right)^{2}\right]$ $=\frac{1}{2} \mathrm{LI}_{\mathrm{o}}^{2} \cdot 2\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right) \cdot\left[-\mathrm{e}^{-\mathrm{t} / \tau} \times \frac{-1}{\tau}\right]$ $\mathrm{P}=\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left(\mathrm{e}^{-\mathrm{t} / \tau}-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)$ Maximum power, $\frac{\mathrm{dP}}{\mathrm{dt}}=0$ $\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{e}^{-\mathrm{t} / \tau}-\mathrm{e}^{-2 \mathrm{t} / \tau}\right)=0$ $\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left[\left(-\frac{1}{\tau}\right) \mathrm{e}^{-\mathrm{t} / \tau}+\frac{2}{\tau} \mathrm{e}^{-2 \mathrm{t} / \tau}\right]=0$ $\frac{1}{\tau} \mathrm{e}^{-\mathrm{t} / \tau}=\frac{2}{\tau} \mathrm{e}^{-2 \mathrm{t} / \tau}$ $\frac{\mathrm{e}^{-2 \mathrm{t} / \tau}}{\mathrm{e}^{-\mathrm{t} / \tau}}=\frac{1}{2}$ $\mathrm{e}^{-\mathrm{t} / \tau}=\frac{1}{2}$ Put the value of $\mathrm{e}^{-\mathrm{t} / \tau}=\frac{1}{2}$ in equation (ii), $\mathrm{P} =\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau}\left[\frac{1}{2}\left(1-\frac{1}{2}\right)\right]$ $\mathrm{P} =\frac{\mathrm{LI}_{\mathrm{o}}^{2}}{\tau} \times \frac{1}{4}$ $=\mathrm{L}\left(\frac{\varepsilon}{\mathrm{R}}\right)^{2} \frac{1}{\mathrm{~L} / \mathrm{R}} \times \frac{1}{4} \quad\left\{\because \mathrm{I}_{\mathrm{o}}=\frac{\varepsilon}{\mathrm{R}}, \tau=\frac{\mathrm{L}}{\mathrm{R}}\right\}$ $\mathrm{P} =\frac{\varepsilon^{2}}{4 \mathrm{R}}$
UPSEE - 2006
Alternating Current
155204
A current of $i=2 \sin (\pi t / 3) \quad A$ is flowing in an inductor of $2 \mathrm{H}$. The amount of work done in increasing the current from $1.0 \mathrm{~A}$ to $2.0 \mathrm{~A}$ is :
1 $1 \mathrm{~J}$
2 $2 \mathrm{~J}$
3 $3 \mathrm{~J}$
4 $4 \mathrm{~J}$
Explanation:
C Given that, Primary current $i_{1}=1 \mathrm{~A}$ Secondary current $\mathrm{i}_{2}=2 \mathrm{~A}$ Self inductance $\mathrm{L}=2 \mathrm{H}$ $\text { Work done }=\text { Change in energy } =\frac{1}{2}\left(\mathrm{Li}_{2}^{2}-\mathrm{Li}_{1}^{2}\right)$ $=\frac{1}{2} \mathrm{~L}\left(\mathrm{i}_{2}^{2}-\mathrm{i}_{1}^{2}\right)$ $=\frac{1}{2} \times 2\left[(2)^{2}-(1)^{2}\right]$ $=4-1$ Work done $=3 \mathrm{~J}$
UPSEE - 2004
Alternating Current
155206
An L-C-R series circuit is at resonance. Then
1 The phase difference between current and voltage is $90^{\circ}$
2 The phase difference between current and voltage is $45^{\circ}$
3 Its impedance is purely resistive
4 Its impedance is zero
5 The current is minimum
Explanation:
C As we know that, at resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ So, impedance $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+0}$ $\mathrm{Z}=\mathrm{R}$ Hence, In an LCR series circuit at resonance impedance is purely resistive.
Kerala CEE - 2010
Alternating Current
155207
An L-C-R series AC circuit is at resonance with $10 \mathrm{~V}$ each across $\mathrm{L}, \mathrm{C}$ and $\mathrm{R}$. If the resistance is halved, the respective voltages across $L, C$ and $R$ are
1 $10 \mathrm{~V}, 10 \mathrm{~V}$ and $5 \mathrm{~V}$
2 $10 \mathrm{~V}, 10 \mathrm{~V}$ and $10 \mathrm{~V}$
3 $20 \mathrm{~V}, 20 \mathrm{~V}$ and $5 \mathrm{~V}$
4 $20 \mathrm{~V}, 20 \mathrm{~V}$ and $10 \mathrm{~V}$
5 $5 \mathrm{~V}, 5 \mathrm{~V}$ and $5 \mathrm{~V}$
Explanation:
D Given, Voltage (v) $=10$ volt Let, Initial current $=\mathrm{i}$ Initial resistance $=\mathrm{R}$ Now, Inductive resistance $=\mathrm{X}_{\mathrm{L}}$ Capacitive resistance $=\mathrm{X}_{\mathrm{C}}$ We know that, When resistance is halved current will be doubled $\therefore \quad \mathrm{v}_{\mathrm{R}}=2 \mathrm{i} \times(\mathrm{R} / 2)=\mathrm{iR}=10 \mathrm{~V}$ $\mathrm{v}_{\mathrm{C}}=2 \mathrm{i} \times \mathrm{X}_{\mathrm{C}}=20 \mathrm{~V}$ $\mathrm{v}_{\mathrm{L}}=2 \mathrm{i} \times \mathrm{X}_{\mathrm{L}}=20 \mathrm{~V}$
