154927
The self inductance of the motor of an electric fan is $10 \mathrm{H}$. In order to impart maximum power at $50 \mathrm{~Hz}$, it should be connected to a capacitance of
1 $1 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
A Given, $\mathrm{L}=10 \mathrm{H}$, frequency, $\mathrm{f}=50 \mathrm{~Hz}$ For maximum power - $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}} \quad(\because \omega=2 \pi \mathrm{f})$ $\mathrm{C}=\frac{1}{\omega^{2} \mathrm{~L}}=0.000001013$ $\mathrm{C}=\frac{1}{(2 \pi \mathrm{f})^{2} \mathrm{~L}}=\frac{1}{4 \pi^{2} \times(50)^{2} \times 10}$ $\mathrm{C}=1 \mu \mathrm{F}$
AMU-2015
Electro Magnetic Induction
154928
Two inductors $0.4 \mathrm{H}$ and $0.6 \mathrm{H}$ are connected in parallel. If this combination is connected in series with an inductor of inductance $0.76 \mathrm{H}$. The equivalent inductance of the circuit will be
1 $0.1 \mathrm{H}$
2 $0.2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $2 \mathrm{H}$
Explanation:
C Given that, Primary inductance, $\mathrm{L}_{1}=0.4 \mathrm{H}$, Secondary Inductance, $\mathrm{L}_{2}=0.6 \mathrm{H}$, $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ are conected parallel to each other - $\mathrm{L}_{\text {eq }}=\frac{\mathrm{L}_{1} \cdot \mathrm{L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ $\mathrm{L}_{\text {eq }}=\frac{0.4 \times 0.6}{0.4+0.6}=0.24 \mathrm{H}$ $\mathrm{L}_{\text {eq }} \text { is connected in series with } \mathrm{L}_{3}$ $\mathrm{~L}=\mathrm{L}_{\mathrm{eq}}+\mathrm{L}_{3}=0.24+0.76=1 \mathrm{H}$
AMU-2001
Electro Magnetic Induction
154929
If the number of turns per unit length of a coil of solenoid is doubled, the self - inductance of the solenoid will
1 remain unchanged
2 be halved
3 be doubled
4 become four times
Explanation:
D We know that, $\mathrm{L} \propto \mathrm{N}^{2}$ $\mathrm{~L}_{1} \propto \mathrm{N}^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2}=\left(\frac{\mathrm{N}_{1}}{2 \mathrm{~N}_{1}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{4}$ $\mathrm{~L}_{2}=4 \mathrm{~L}_{1}$ So, if $\mathrm{N}$ becomes double then $\mathrm{L}$ will become four times.
AIPMT- 1991
Electro Magnetic Induction
154930
The current (I) in the inductance is varying with time according to the plot shown in figure. Which one of the following is the correct variation of voltage with time in the coil?
1
2
3
4
Explanation:
D Since current is linearly increasing the voltage is a linerly constant, when current is linearly decreasing voltage is negative constant. $\mathrm{E}=\frac{-\mathrm{Ldi}}{\mathrm{dt}}$ $|\mathrm{E}| \propto \frac{\mathrm{dI}}{\mathrm{dt}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154927
The self inductance of the motor of an electric fan is $10 \mathrm{H}$. In order to impart maximum power at $50 \mathrm{~Hz}$, it should be connected to a capacitance of
1 $1 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
A Given, $\mathrm{L}=10 \mathrm{H}$, frequency, $\mathrm{f}=50 \mathrm{~Hz}$ For maximum power - $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}} \quad(\because \omega=2 \pi \mathrm{f})$ $\mathrm{C}=\frac{1}{\omega^{2} \mathrm{~L}}=0.000001013$ $\mathrm{C}=\frac{1}{(2 \pi \mathrm{f})^{2} \mathrm{~L}}=\frac{1}{4 \pi^{2} \times(50)^{2} \times 10}$ $\mathrm{C}=1 \mu \mathrm{F}$
AMU-2015
Electro Magnetic Induction
154928
Two inductors $0.4 \mathrm{H}$ and $0.6 \mathrm{H}$ are connected in parallel. If this combination is connected in series with an inductor of inductance $0.76 \mathrm{H}$. The equivalent inductance of the circuit will be
1 $0.1 \mathrm{H}$
2 $0.2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $2 \mathrm{H}$
Explanation:
C Given that, Primary inductance, $\mathrm{L}_{1}=0.4 \mathrm{H}$, Secondary Inductance, $\mathrm{L}_{2}=0.6 \mathrm{H}$, $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ are conected parallel to each other - $\mathrm{L}_{\text {eq }}=\frac{\mathrm{L}_{1} \cdot \mathrm{L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ $\mathrm{L}_{\text {eq }}=\frac{0.4 \times 0.6}{0.4+0.6}=0.24 \mathrm{H}$ $\mathrm{L}_{\text {eq }} \text { is connected in series with } \mathrm{L}_{3}$ $\mathrm{~L}=\mathrm{L}_{\mathrm{eq}}+\mathrm{L}_{3}=0.24+0.76=1 \mathrm{H}$
AMU-2001
Electro Magnetic Induction
154929
If the number of turns per unit length of a coil of solenoid is doubled, the self - inductance of the solenoid will
1 remain unchanged
2 be halved
3 be doubled
4 become four times
Explanation:
D We know that, $\mathrm{L} \propto \mathrm{N}^{2}$ $\mathrm{~L}_{1} \propto \mathrm{N}^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2}=\left(\frac{\mathrm{N}_{1}}{2 \mathrm{~N}_{1}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{4}$ $\mathrm{~L}_{2}=4 \mathrm{~L}_{1}$ So, if $\mathrm{N}$ becomes double then $\mathrm{L}$ will become four times.
AIPMT- 1991
Electro Magnetic Induction
154930
The current (I) in the inductance is varying with time according to the plot shown in figure. Which one of the following is the correct variation of voltage with time in the coil?
1
2
3
4
Explanation:
D Since current is linearly increasing the voltage is a linerly constant, when current is linearly decreasing voltage is negative constant. $\mathrm{E}=\frac{-\mathrm{Ldi}}{\mathrm{dt}}$ $|\mathrm{E}| \propto \frac{\mathrm{dI}}{\mathrm{dt}}$
154927
The self inductance of the motor of an electric fan is $10 \mathrm{H}$. In order to impart maximum power at $50 \mathrm{~Hz}$, it should be connected to a capacitance of
1 $1 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
A Given, $\mathrm{L}=10 \mathrm{H}$, frequency, $\mathrm{f}=50 \mathrm{~Hz}$ For maximum power - $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}} \quad(\because \omega=2 \pi \mathrm{f})$ $\mathrm{C}=\frac{1}{\omega^{2} \mathrm{~L}}=0.000001013$ $\mathrm{C}=\frac{1}{(2 \pi \mathrm{f})^{2} \mathrm{~L}}=\frac{1}{4 \pi^{2} \times(50)^{2} \times 10}$ $\mathrm{C}=1 \mu \mathrm{F}$
AMU-2015
Electro Magnetic Induction
154928
Two inductors $0.4 \mathrm{H}$ and $0.6 \mathrm{H}$ are connected in parallel. If this combination is connected in series with an inductor of inductance $0.76 \mathrm{H}$. The equivalent inductance of the circuit will be
1 $0.1 \mathrm{H}$
2 $0.2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $2 \mathrm{H}$
Explanation:
C Given that, Primary inductance, $\mathrm{L}_{1}=0.4 \mathrm{H}$, Secondary Inductance, $\mathrm{L}_{2}=0.6 \mathrm{H}$, $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ are conected parallel to each other - $\mathrm{L}_{\text {eq }}=\frac{\mathrm{L}_{1} \cdot \mathrm{L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ $\mathrm{L}_{\text {eq }}=\frac{0.4 \times 0.6}{0.4+0.6}=0.24 \mathrm{H}$ $\mathrm{L}_{\text {eq }} \text { is connected in series with } \mathrm{L}_{3}$ $\mathrm{~L}=\mathrm{L}_{\mathrm{eq}}+\mathrm{L}_{3}=0.24+0.76=1 \mathrm{H}$
AMU-2001
Electro Magnetic Induction
154929
If the number of turns per unit length of a coil of solenoid is doubled, the self - inductance of the solenoid will
1 remain unchanged
2 be halved
3 be doubled
4 become four times
Explanation:
D We know that, $\mathrm{L} \propto \mathrm{N}^{2}$ $\mathrm{~L}_{1} \propto \mathrm{N}^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2}=\left(\frac{\mathrm{N}_{1}}{2 \mathrm{~N}_{1}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{4}$ $\mathrm{~L}_{2}=4 \mathrm{~L}_{1}$ So, if $\mathrm{N}$ becomes double then $\mathrm{L}$ will become four times.
AIPMT- 1991
Electro Magnetic Induction
154930
The current (I) in the inductance is varying with time according to the plot shown in figure. Which one of the following is the correct variation of voltage with time in the coil?
1
2
3
4
Explanation:
D Since current is linearly increasing the voltage is a linerly constant, when current is linearly decreasing voltage is negative constant. $\mathrm{E}=\frac{-\mathrm{Ldi}}{\mathrm{dt}}$ $|\mathrm{E}| \propto \frac{\mathrm{dI}}{\mathrm{dt}}$
154927
The self inductance of the motor of an electric fan is $10 \mathrm{H}$. In order to impart maximum power at $50 \mathrm{~Hz}$, it should be connected to a capacitance of
1 $1 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
A Given, $\mathrm{L}=10 \mathrm{H}$, frequency, $\mathrm{f}=50 \mathrm{~Hz}$ For maximum power - $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}} \quad(\because \omega=2 \pi \mathrm{f})$ $\mathrm{C}=\frac{1}{\omega^{2} \mathrm{~L}}=0.000001013$ $\mathrm{C}=\frac{1}{(2 \pi \mathrm{f})^{2} \mathrm{~L}}=\frac{1}{4 \pi^{2} \times(50)^{2} \times 10}$ $\mathrm{C}=1 \mu \mathrm{F}$
AMU-2015
Electro Magnetic Induction
154928
Two inductors $0.4 \mathrm{H}$ and $0.6 \mathrm{H}$ are connected in parallel. If this combination is connected in series with an inductor of inductance $0.76 \mathrm{H}$. The equivalent inductance of the circuit will be
1 $0.1 \mathrm{H}$
2 $0.2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $2 \mathrm{H}$
Explanation:
C Given that, Primary inductance, $\mathrm{L}_{1}=0.4 \mathrm{H}$, Secondary Inductance, $\mathrm{L}_{2}=0.6 \mathrm{H}$, $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ are conected parallel to each other - $\mathrm{L}_{\text {eq }}=\frac{\mathrm{L}_{1} \cdot \mathrm{L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ $\mathrm{L}_{\text {eq }}=\frac{0.4 \times 0.6}{0.4+0.6}=0.24 \mathrm{H}$ $\mathrm{L}_{\text {eq }} \text { is connected in series with } \mathrm{L}_{3}$ $\mathrm{~L}=\mathrm{L}_{\mathrm{eq}}+\mathrm{L}_{3}=0.24+0.76=1 \mathrm{H}$
AMU-2001
Electro Magnetic Induction
154929
If the number of turns per unit length of a coil of solenoid is doubled, the self - inductance of the solenoid will
1 remain unchanged
2 be halved
3 be doubled
4 become four times
Explanation:
D We know that, $\mathrm{L} \propto \mathrm{N}^{2}$ $\mathrm{~L}_{1} \propto \mathrm{N}^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\left(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}\right)^{2}=\left(\frac{\mathrm{N}_{1}}{2 \mathrm{~N}_{1}}\right)^{2}$ $\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}=\frac{1}{4}$ $\mathrm{~L}_{2}=4 \mathrm{~L}_{1}$ So, if $\mathrm{N}$ becomes double then $\mathrm{L}$ will become four times.
AIPMT- 1991
Electro Magnetic Induction
154930
The current (I) in the inductance is varying with time according to the plot shown in figure. Which one of the following is the correct variation of voltage with time in the coil?
1
2
3
4
Explanation:
D Since current is linearly increasing the voltage is a linerly constant, when current is linearly decreasing voltage is negative constant. $\mathrm{E}=\frac{-\mathrm{Ldi}}{\mathrm{dt}}$ $|\mathrm{E}| \propto \frac{\mathrm{dI}}{\mathrm{dt}}$
