154891
An electric bulb in series with a large inductor when connected across a D.C. source takes a little time before reaching a stable glow. If an iron core is inserted into the inductor the time delay of reaching stable glow will
1 decrease
2 increase
3 remains the same
4 may increase or decrease
Explanation:
B $I=\frac{V}{\sqrt{X^{2}{ }_{L}+R^{2}}}$ $X_{L}=\omega L$ When we use soft iron rod into the inductor Then inductance $\mathrm{L}$ increase. So, $X_{\mathrm{L}}=\omega \mathrm{L}$ is also increase So, $\mathrm{Z}$ increase Hence, current through Bulb will decrease so the time delay will increase to reaching stable glow-
J and K CET- 2001
Electro Magnetic Induction
154893
When the number of turns in a coil is doubled without any change in the length of the coil, its self inductance becomes
1 4 times
2 2 times
3 halved
4 remains unchanged
Explanation:
A Self inductance $\mathrm{L}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ $\mathrm{~L} \propto \mathrm{N}^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{2 \mathrm{~N}_{1}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{4} \Rightarrow \mathrm{L}_{2}=4 \mathrm{~L}_{1}$ $\therefore$ Inductance become 4 times
J and K CET- 2000
Electro Magnetic Induction
154894
The current in a coil changes from $4 \mathrm{~A}$ to zero in $0.1 \mathrm{~s}$. If average e.m.f. induced is $100 \mathrm{~V}$, what is self-inductance of coil?
1 $2.5 \mathrm{H}$
2 $25 \mathrm{H}$
3 $400 \mathrm{H}$
4 $40 \mathrm{H}$
Explanation:
A change in current $\Delta \mathrm{I}=4-0=4 \mathrm{~A}$ Time taken $\Delta \mathrm{t}=0.1 \mathrm{sec}$ emf induced $\varepsilon=100 \mathrm{~V}$ $\varepsilon=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\mathrm{L}=\varepsilon \frac{\mathrm{dt}}{\mathrm{di}}=\frac{100 \times 0.1}{4}$ $\mathrm{L}=25 \times .1=2.5$ $\mathrm{L}=2.5 \mathrm{H}$
J and K CET- 1999
Electro Magnetic Induction
154896
An e.m.f. of 8 volt is induced in a coil when current in it rises from $2 \mathrm{~A}$ to $4 \mathrm{~A}$ in 0.05 sec. The self inductance of the coil is
1 $0.1 \mathrm{H}$
2 $0.2 \mathrm{H}$
3 $0.4 \mathrm{H}$
4 $0.8 \mathrm{H}$
Explanation:
B : Given emf $\varepsilon=8 \mathrm{v}$ Charge in current $=4-2=2 \mathrm{~A}$ Time $\mathrm{dt}=0.05 \mathrm{sec}$ $\varepsilon=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\mathrm{L}=\varepsilon \frac{\mathrm{dt}}{\mathrm{di}}=\frac{8 \times 0.05}{2}$ $\mathrm{~L}=4 \times 0.05$ $\mathrm{~L}=0.20 \mathrm{H}$
154891
An electric bulb in series with a large inductor when connected across a D.C. source takes a little time before reaching a stable glow. If an iron core is inserted into the inductor the time delay of reaching stable glow will
1 decrease
2 increase
3 remains the same
4 may increase or decrease
Explanation:
B $I=\frac{V}{\sqrt{X^{2}{ }_{L}+R^{2}}}$ $X_{L}=\omega L$ When we use soft iron rod into the inductor Then inductance $\mathrm{L}$ increase. So, $X_{\mathrm{L}}=\omega \mathrm{L}$ is also increase So, $\mathrm{Z}$ increase Hence, current through Bulb will decrease so the time delay will increase to reaching stable glow-
J and K CET- 2001
Electro Magnetic Induction
154893
When the number of turns in a coil is doubled without any change in the length of the coil, its self inductance becomes
1 4 times
2 2 times
3 halved
4 remains unchanged
Explanation:
A Self inductance $\mathrm{L}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ $\mathrm{~L} \propto \mathrm{N}^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{2 \mathrm{~N}_{1}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{4} \Rightarrow \mathrm{L}_{2}=4 \mathrm{~L}_{1}$ $\therefore$ Inductance become 4 times
J and K CET- 2000
Electro Magnetic Induction
154894
The current in a coil changes from $4 \mathrm{~A}$ to zero in $0.1 \mathrm{~s}$. If average e.m.f. induced is $100 \mathrm{~V}$, what is self-inductance of coil?
1 $2.5 \mathrm{H}$
2 $25 \mathrm{H}$
3 $400 \mathrm{H}$
4 $40 \mathrm{H}$
Explanation:
A change in current $\Delta \mathrm{I}=4-0=4 \mathrm{~A}$ Time taken $\Delta \mathrm{t}=0.1 \mathrm{sec}$ emf induced $\varepsilon=100 \mathrm{~V}$ $\varepsilon=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\mathrm{L}=\varepsilon \frac{\mathrm{dt}}{\mathrm{di}}=\frac{100 \times 0.1}{4}$ $\mathrm{L}=25 \times .1=2.5$ $\mathrm{L}=2.5 \mathrm{H}$
J and K CET- 1999
Electro Magnetic Induction
154896
An e.m.f. of 8 volt is induced in a coil when current in it rises from $2 \mathrm{~A}$ to $4 \mathrm{~A}$ in 0.05 sec. The self inductance of the coil is
1 $0.1 \mathrm{H}$
2 $0.2 \mathrm{H}$
3 $0.4 \mathrm{H}$
4 $0.8 \mathrm{H}$
Explanation:
B : Given emf $\varepsilon=8 \mathrm{v}$ Charge in current $=4-2=2 \mathrm{~A}$ Time $\mathrm{dt}=0.05 \mathrm{sec}$ $\varepsilon=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\mathrm{L}=\varepsilon \frac{\mathrm{dt}}{\mathrm{di}}=\frac{8 \times 0.05}{2}$ $\mathrm{~L}=4 \times 0.05$ $\mathrm{~L}=0.20 \mathrm{H}$
154891
An electric bulb in series with a large inductor when connected across a D.C. source takes a little time before reaching a stable glow. If an iron core is inserted into the inductor the time delay of reaching stable glow will
1 decrease
2 increase
3 remains the same
4 may increase or decrease
Explanation:
B $I=\frac{V}{\sqrt{X^{2}{ }_{L}+R^{2}}}$ $X_{L}=\omega L$ When we use soft iron rod into the inductor Then inductance $\mathrm{L}$ increase. So, $X_{\mathrm{L}}=\omega \mathrm{L}$ is also increase So, $\mathrm{Z}$ increase Hence, current through Bulb will decrease so the time delay will increase to reaching stable glow-
J and K CET- 2001
Electro Magnetic Induction
154893
When the number of turns in a coil is doubled without any change in the length of the coil, its self inductance becomes
1 4 times
2 2 times
3 halved
4 remains unchanged
Explanation:
A Self inductance $\mathrm{L}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ $\mathrm{~L} \propto \mathrm{N}^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{2 \mathrm{~N}_{1}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{4} \Rightarrow \mathrm{L}_{2}=4 \mathrm{~L}_{1}$ $\therefore$ Inductance become 4 times
J and K CET- 2000
Electro Magnetic Induction
154894
The current in a coil changes from $4 \mathrm{~A}$ to zero in $0.1 \mathrm{~s}$. If average e.m.f. induced is $100 \mathrm{~V}$, what is self-inductance of coil?
1 $2.5 \mathrm{H}$
2 $25 \mathrm{H}$
3 $400 \mathrm{H}$
4 $40 \mathrm{H}$
Explanation:
A change in current $\Delta \mathrm{I}=4-0=4 \mathrm{~A}$ Time taken $\Delta \mathrm{t}=0.1 \mathrm{sec}$ emf induced $\varepsilon=100 \mathrm{~V}$ $\varepsilon=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\mathrm{L}=\varepsilon \frac{\mathrm{dt}}{\mathrm{di}}=\frac{100 \times 0.1}{4}$ $\mathrm{L}=25 \times .1=2.5$ $\mathrm{L}=2.5 \mathrm{H}$
J and K CET- 1999
Electro Magnetic Induction
154896
An e.m.f. of 8 volt is induced in a coil when current in it rises from $2 \mathrm{~A}$ to $4 \mathrm{~A}$ in 0.05 sec. The self inductance of the coil is
1 $0.1 \mathrm{H}$
2 $0.2 \mathrm{H}$
3 $0.4 \mathrm{H}$
4 $0.8 \mathrm{H}$
Explanation:
B : Given emf $\varepsilon=8 \mathrm{v}$ Charge in current $=4-2=2 \mathrm{~A}$ Time $\mathrm{dt}=0.05 \mathrm{sec}$ $\varepsilon=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\mathrm{L}=\varepsilon \frac{\mathrm{dt}}{\mathrm{di}}=\frac{8 \times 0.05}{2}$ $\mathrm{~L}=4 \times 0.05$ $\mathrm{~L}=0.20 \mathrm{H}$
154891
An electric bulb in series with a large inductor when connected across a D.C. source takes a little time before reaching a stable glow. If an iron core is inserted into the inductor the time delay of reaching stable glow will
1 decrease
2 increase
3 remains the same
4 may increase or decrease
Explanation:
B $I=\frac{V}{\sqrt{X^{2}{ }_{L}+R^{2}}}$ $X_{L}=\omega L$ When we use soft iron rod into the inductor Then inductance $\mathrm{L}$ increase. So, $X_{\mathrm{L}}=\omega \mathrm{L}$ is also increase So, $\mathrm{Z}$ increase Hence, current through Bulb will decrease so the time delay will increase to reaching stable glow-
J and K CET- 2001
Electro Magnetic Induction
154893
When the number of turns in a coil is doubled without any change in the length of the coil, its self inductance becomes
1 4 times
2 2 times
3 halved
4 remains unchanged
Explanation:
A Self inductance $\mathrm{L}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ $\mathrm{~L} \propto \mathrm{N}^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{2 \mathrm{~N}_{1}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{4} \Rightarrow \mathrm{L}_{2}=4 \mathrm{~L}_{1}$ $\therefore$ Inductance become 4 times
J and K CET- 2000
Electro Magnetic Induction
154894
The current in a coil changes from $4 \mathrm{~A}$ to zero in $0.1 \mathrm{~s}$. If average e.m.f. induced is $100 \mathrm{~V}$, what is self-inductance of coil?
1 $2.5 \mathrm{H}$
2 $25 \mathrm{H}$
3 $400 \mathrm{H}$
4 $40 \mathrm{H}$
Explanation:
A change in current $\Delta \mathrm{I}=4-0=4 \mathrm{~A}$ Time taken $\Delta \mathrm{t}=0.1 \mathrm{sec}$ emf induced $\varepsilon=100 \mathrm{~V}$ $\varepsilon=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\mathrm{L}=\varepsilon \frac{\mathrm{dt}}{\mathrm{di}}=\frac{100 \times 0.1}{4}$ $\mathrm{L}=25 \times .1=2.5$ $\mathrm{L}=2.5 \mathrm{H}$
J and K CET- 1999
Electro Magnetic Induction
154896
An e.m.f. of 8 volt is induced in a coil when current in it rises from $2 \mathrm{~A}$ to $4 \mathrm{~A}$ in 0.05 sec. The self inductance of the coil is
1 $0.1 \mathrm{H}$
2 $0.2 \mathrm{H}$
3 $0.4 \mathrm{H}$
4 $0.8 \mathrm{H}$
Explanation:
B : Given emf $\varepsilon=8 \mathrm{v}$ Charge in current $=4-2=2 \mathrm{~A}$ Time $\mathrm{dt}=0.05 \mathrm{sec}$ $\varepsilon=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\mathrm{L}=\varepsilon \frac{\mathrm{dt}}{\mathrm{di}}=\frac{8 \times 0.05}{2}$ $\mathrm{~L}=4 \times 0.05$ $\mathrm{~L}=0.20 \mathrm{H}$
