154918
In $0.2 \mathrm{~s}$, the current in a coil increases from 2.0 A to 3.0 A. If inductance of coil is $60 \mathrm{mH}$, then induced current in external resistance of $3 \Omega$ will be
1 $1 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.1 \mathrm{~A}$
Explanation:
D Given, Change in current $(\Delta \mathrm{I})=3-2=1 \mathrm{~A}$ Inductance of coil $(\mathrm{L})=60 \times 10^{-3} \mathrm{H}$ Time $(\Delta \mathrm{t})=0.2 \mathrm{~s}$ Induced emf $\varepsilon=\mathrm{L} \cdot \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}=60 \times 10^{-3} \times \frac{1}{0.2}=300 \times 10^{-3}$ $\varepsilon=3 \times 10^{-1} \mathrm{~V}$ Induced current $(\mathrm{I})=\frac{\varepsilon}{\mathrm{R}}=\frac{3 \times 10^{-1}}{3}$ $\mathrm{I}=1 \times 10^{-1}=0.1 \mathrm{~A}$ $\mathrm{I}=0.1 \mathrm{~A}$
JIPMER-2013
Electro Magnetic Induction
154919
$220 \mathrm{~V}, 50 \mathrm{~Hz}, \mathrm{AC}$ source is connected to an inductance of $0.2 \mathrm{H}$ and a resistance of $20 \Omega$ in series. What is the current in the circuit?
1 $3.33 \mathrm{~A}$
2 $33.3 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
A Given, Voltage $(\mathrm{V})=220$ Vo, Frequency $(\mathrm{f})=50 \mathrm{~Hz}$, Inductancen $(A)=0.2 \mathrm{H}$ and Resistance $(\mathrm{R})=20 \Omega$ Inductive Reactance $\mathrm{X}_{\mathrm{L}} =\omega \mathrm{L}=2 \pi \mathrm{f} \mathrm{L}$ $=2 \pi \times 50 \times 0.2$ $\mathrm{X}_{\mathrm{L}} =20 \pi \Omega$ Impedance $Z=\sqrt{X_{L}^{2}+R^{2}}$ $Z=\sqrt{(20)^{2}+(20 \pi)^{2}}$ Current $I=\frac{V}{Z}=\frac{220}{20 \sqrt{1+\pi^{2}}}$ $\mathrm{I}=3.33 \mathrm{~A}$
JIPMER-2013
Electro Magnetic Induction
154920
In $0.1 \mathrm{~s}$, the current in a coil increases from $1 \mathrm{~A}$ to 1.5A. If inductance of coil is $60 \mathrm{mH}$, then induced current in external resistance of $3 \Omega$ will be
1 $1 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.1 \mathrm{~A}$
Explanation:
D Givne, Change in current $(\Delta \mathrm{I})=(1.5-1)$ $\Delta \mathrm{I}=0.5 \mathrm{~A}$ Inductance of the coil $(\mathrm{L})=60 \mathrm{mH}=60 \times 10^{-3} \mathrm{H}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ Induced emf $\varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}=60 \times 10^{-3} \times \frac{0.5}{0.1}$ $\varepsilon=0.3 \mathrm{~V}$ Current in external Resistance $I=\frac{\varepsilon}{R}=\frac{0.3}{3}=0.1$ $I=0.1 \mathrm{~A}$
JIPMER-2010
Electro Magnetic Induction
154921
If the current 3.0A flowing in the primary coil is made zero in $0.1 \mathrm{sec}$, the emf induced in the secondary coil is 1.5 volt. The mutual inductance between the coils is
1 $0.05 \mathrm{H}$
2 $1.05 \mathrm{H}$
3 $0.1 \mathrm{H}$
4 $0.2 \mathrm{H}$
Explanation:
A Given, Change in current $(\Delta \mathrm{I})=(0-3)=-3 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ emf induced $(\varepsilon)=1.5 \mathrm{~V}$ $\varepsilon=M \cdot \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $1.5=\frac{\mathrm{M} \times 3}{0.1}$ $\mathrm{M}=0.05 \mathrm{H}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154918
In $0.2 \mathrm{~s}$, the current in a coil increases from 2.0 A to 3.0 A. If inductance of coil is $60 \mathrm{mH}$, then induced current in external resistance of $3 \Omega$ will be
1 $1 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.1 \mathrm{~A}$
Explanation:
D Given, Change in current $(\Delta \mathrm{I})=3-2=1 \mathrm{~A}$ Inductance of coil $(\mathrm{L})=60 \times 10^{-3} \mathrm{H}$ Time $(\Delta \mathrm{t})=0.2 \mathrm{~s}$ Induced emf $\varepsilon=\mathrm{L} \cdot \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}=60 \times 10^{-3} \times \frac{1}{0.2}=300 \times 10^{-3}$ $\varepsilon=3 \times 10^{-1} \mathrm{~V}$ Induced current $(\mathrm{I})=\frac{\varepsilon}{\mathrm{R}}=\frac{3 \times 10^{-1}}{3}$ $\mathrm{I}=1 \times 10^{-1}=0.1 \mathrm{~A}$ $\mathrm{I}=0.1 \mathrm{~A}$
JIPMER-2013
Electro Magnetic Induction
154919
$220 \mathrm{~V}, 50 \mathrm{~Hz}, \mathrm{AC}$ source is connected to an inductance of $0.2 \mathrm{H}$ and a resistance of $20 \Omega$ in series. What is the current in the circuit?
1 $3.33 \mathrm{~A}$
2 $33.3 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
A Given, Voltage $(\mathrm{V})=220$ Vo, Frequency $(\mathrm{f})=50 \mathrm{~Hz}$, Inductancen $(A)=0.2 \mathrm{H}$ and Resistance $(\mathrm{R})=20 \Omega$ Inductive Reactance $\mathrm{X}_{\mathrm{L}} =\omega \mathrm{L}=2 \pi \mathrm{f} \mathrm{L}$ $=2 \pi \times 50 \times 0.2$ $\mathrm{X}_{\mathrm{L}} =20 \pi \Omega$ Impedance $Z=\sqrt{X_{L}^{2}+R^{2}}$ $Z=\sqrt{(20)^{2}+(20 \pi)^{2}}$ Current $I=\frac{V}{Z}=\frac{220}{20 \sqrt{1+\pi^{2}}}$ $\mathrm{I}=3.33 \mathrm{~A}$
JIPMER-2013
Electro Magnetic Induction
154920
In $0.1 \mathrm{~s}$, the current in a coil increases from $1 \mathrm{~A}$ to 1.5A. If inductance of coil is $60 \mathrm{mH}$, then induced current in external resistance of $3 \Omega$ will be
1 $1 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.1 \mathrm{~A}$
Explanation:
D Givne, Change in current $(\Delta \mathrm{I})=(1.5-1)$ $\Delta \mathrm{I}=0.5 \mathrm{~A}$ Inductance of the coil $(\mathrm{L})=60 \mathrm{mH}=60 \times 10^{-3} \mathrm{H}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ Induced emf $\varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}=60 \times 10^{-3} \times \frac{0.5}{0.1}$ $\varepsilon=0.3 \mathrm{~V}$ Current in external Resistance $I=\frac{\varepsilon}{R}=\frac{0.3}{3}=0.1$ $I=0.1 \mathrm{~A}$
JIPMER-2010
Electro Magnetic Induction
154921
If the current 3.0A flowing in the primary coil is made zero in $0.1 \mathrm{sec}$, the emf induced in the secondary coil is 1.5 volt. The mutual inductance between the coils is
1 $0.05 \mathrm{H}$
2 $1.05 \mathrm{H}$
3 $0.1 \mathrm{H}$
4 $0.2 \mathrm{H}$
Explanation:
A Given, Change in current $(\Delta \mathrm{I})=(0-3)=-3 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ emf induced $(\varepsilon)=1.5 \mathrm{~V}$ $\varepsilon=M \cdot \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $1.5=\frac{\mathrm{M} \times 3}{0.1}$ $\mathrm{M}=0.05 \mathrm{H}$
154918
In $0.2 \mathrm{~s}$, the current in a coil increases from 2.0 A to 3.0 A. If inductance of coil is $60 \mathrm{mH}$, then induced current in external resistance of $3 \Omega$ will be
1 $1 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.1 \mathrm{~A}$
Explanation:
D Given, Change in current $(\Delta \mathrm{I})=3-2=1 \mathrm{~A}$ Inductance of coil $(\mathrm{L})=60 \times 10^{-3} \mathrm{H}$ Time $(\Delta \mathrm{t})=0.2 \mathrm{~s}$ Induced emf $\varepsilon=\mathrm{L} \cdot \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}=60 \times 10^{-3} \times \frac{1}{0.2}=300 \times 10^{-3}$ $\varepsilon=3 \times 10^{-1} \mathrm{~V}$ Induced current $(\mathrm{I})=\frac{\varepsilon}{\mathrm{R}}=\frac{3 \times 10^{-1}}{3}$ $\mathrm{I}=1 \times 10^{-1}=0.1 \mathrm{~A}$ $\mathrm{I}=0.1 \mathrm{~A}$
JIPMER-2013
Electro Magnetic Induction
154919
$220 \mathrm{~V}, 50 \mathrm{~Hz}, \mathrm{AC}$ source is connected to an inductance of $0.2 \mathrm{H}$ and a resistance of $20 \Omega$ in series. What is the current in the circuit?
1 $3.33 \mathrm{~A}$
2 $33.3 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
A Given, Voltage $(\mathrm{V})=220$ Vo, Frequency $(\mathrm{f})=50 \mathrm{~Hz}$, Inductancen $(A)=0.2 \mathrm{H}$ and Resistance $(\mathrm{R})=20 \Omega$ Inductive Reactance $\mathrm{X}_{\mathrm{L}} =\omega \mathrm{L}=2 \pi \mathrm{f} \mathrm{L}$ $=2 \pi \times 50 \times 0.2$ $\mathrm{X}_{\mathrm{L}} =20 \pi \Omega$ Impedance $Z=\sqrt{X_{L}^{2}+R^{2}}$ $Z=\sqrt{(20)^{2}+(20 \pi)^{2}}$ Current $I=\frac{V}{Z}=\frac{220}{20 \sqrt{1+\pi^{2}}}$ $\mathrm{I}=3.33 \mathrm{~A}$
JIPMER-2013
Electro Magnetic Induction
154920
In $0.1 \mathrm{~s}$, the current in a coil increases from $1 \mathrm{~A}$ to 1.5A. If inductance of coil is $60 \mathrm{mH}$, then induced current in external resistance of $3 \Omega$ will be
1 $1 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.1 \mathrm{~A}$
Explanation:
D Givne, Change in current $(\Delta \mathrm{I})=(1.5-1)$ $\Delta \mathrm{I}=0.5 \mathrm{~A}$ Inductance of the coil $(\mathrm{L})=60 \mathrm{mH}=60 \times 10^{-3} \mathrm{H}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ Induced emf $\varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}=60 \times 10^{-3} \times \frac{0.5}{0.1}$ $\varepsilon=0.3 \mathrm{~V}$ Current in external Resistance $I=\frac{\varepsilon}{R}=\frac{0.3}{3}=0.1$ $I=0.1 \mathrm{~A}$
JIPMER-2010
Electro Magnetic Induction
154921
If the current 3.0A flowing in the primary coil is made zero in $0.1 \mathrm{sec}$, the emf induced in the secondary coil is 1.5 volt. The mutual inductance between the coils is
1 $0.05 \mathrm{H}$
2 $1.05 \mathrm{H}$
3 $0.1 \mathrm{H}$
4 $0.2 \mathrm{H}$
Explanation:
A Given, Change in current $(\Delta \mathrm{I})=(0-3)=-3 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ emf induced $(\varepsilon)=1.5 \mathrm{~V}$ $\varepsilon=M \cdot \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $1.5=\frac{\mathrm{M} \times 3}{0.1}$ $\mathrm{M}=0.05 \mathrm{H}$
154918
In $0.2 \mathrm{~s}$, the current in a coil increases from 2.0 A to 3.0 A. If inductance of coil is $60 \mathrm{mH}$, then induced current in external resistance of $3 \Omega$ will be
1 $1 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.1 \mathrm{~A}$
Explanation:
D Given, Change in current $(\Delta \mathrm{I})=3-2=1 \mathrm{~A}$ Inductance of coil $(\mathrm{L})=60 \times 10^{-3} \mathrm{H}$ Time $(\Delta \mathrm{t})=0.2 \mathrm{~s}$ Induced emf $\varepsilon=\mathrm{L} \cdot \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}=60 \times 10^{-3} \times \frac{1}{0.2}=300 \times 10^{-3}$ $\varepsilon=3 \times 10^{-1} \mathrm{~V}$ Induced current $(\mathrm{I})=\frac{\varepsilon}{\mathrm{R}}=\frac{3 \times 10^{-1}}{3}$ $\mathrm{I}=1 \times 10^{-1}=0.1 \mathrm{~A}$ $\mathrm{I}=0.1 \mathrm{~A}$
JIPMER-2013
Electro Magnetic Induction
154919
$220 \mathrm{~V}, 50 \mathrm{~Hz}, \mathrm{AC}$ source is connected to an inductance of $0.2 \mathrm{H}$ and a resistance of $20 \Omega$ in series. What is the current in the circuit?
1 $3.33 \mathrm{~A}$
2 $33.3 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
A Given, Voltage $(\mathrm{V})=220$ Vo, Frequency $(\mathrm{f})=50 \mathrm{~Hz}$, Inductancen $(A)=0.2 \mathrm{H}$ and Resistance $(\mathrm{R})=20 \Omega$ Inductive Reactance $\mathrm{X}_{\mathrm{L}} =\omega \mathrm{L}=2 \pi \mathrm{f} \mathrm{L}$ $=2 \pi \times 50 \times 0.2$ $\mathrm{X}_{\mathrm{L}} =20 \pi \Omega$ Impedance $Z=\sqrt{X_{L}^{2}+R^{2}}$ $Z=\sqrt{(20)^{2}+(20 \pi)^{2}}$ Current $I=\frac{V}{Z}=\frac{220}{20 \sqrt{1+\pi^{2}}}$ $\mathrm{I}=3.33 \mathrm{~A}$
JIPMER-2013
Electro Magnetic Induction
154920
In $0.1 \mathrm{~s}$, the current in a coil increases from $1 \mathrm{~A}$ to 1.5A. If inductance of coil is $60 \mathrm{mH}$, then induced current in external resistance of $3 \Omega$ will be
1 $1 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $0.2 \mathrm{~A}$
4 $0.1 \mathrm{~A}$
Explanation:
D Givne, Change in current $(\Delta \mathrm{I})=(1.5-1)$ $\Delta \mathrm{I}=0.5 \mathrm{~A}$ Inductance of the coil $(\mathrm{L})=60 \mathrm{mH}=60 \times 10^{-3} \mathrm{H}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ Induced emf $\varepsilon=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}=60 \times 10^{-3} \times \frac{0.5}{0.1}$ $\varepsilon=0.3 \mathrm{~V}$ Current in external Resistance $I=\frac{\varepsilon}{R}=\frac{0.3}{3}=0.1$ $I=0.1 \mathrm{~A}$
JIPMER-2010
Electro Magnetic Induction
154921
If the current 3.0A flowing in the primary coil is made zero in $0.1 \mathrm{sec}$, the emf induced in the secondary coil is 1.5 volt. The mutual inductance between the coils is
1 $0.05 \mathrm{H}$
2 $1.05 \mathrm{H}$
3 $0.1 \mathrm{H}$
4 $0.2 \mathrm{H}$
Explanation:
A Given, Change in current $(\Delta \mathrm{I})=(0-3)=-3 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ emf induced $(\varepsilon)=1.5 \mathrm{~V}$ $\varepsilon=M \cdot \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $1.5=\frac{\mathrm{M} \times 3}{0.1}$ $\mathrm{M}=0.05 \mathrm{H}$
