155023
A $0.01 \mathrm{H}$ inductor and $\sqrt{3} \pi \mathrm{ohm}$ resistance are connected in series with a $220 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}$ source. The phase difference between the current and emf is
1 $\frac{\pi}{2} \mathrm{rad}$
2 $\frac{\pi}{6} \mathrm{rad}$
3 $\frac{\pi}{3} \mathrm{rad}$
4 $\frac{\pi}{4} \mathrm{rad}$
Explanation:
B Given, Inductor $(\mathrm{L})=0.01 \mathrm{H}$ Resistance $(\mathrm{R})=\sqrt{3} \pi \Omega$ $\because$ Inductor and resistance are connected in series Then, Phase difference $(\phi)=\tan ^{-1} \frac{X_{L}}{R}$ $\phi=\tan ^{-1}\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)$ $\phi=\tan ^{-1}\left(\frac{2 \pi \mathrm{fL}}{\mathrm{R}}\right)$ Putting all the value in equation (i), we get - $\phi=\tan ^{-1}\left(\frac{2 \pi \times 50 \times 0.01}{\sqrt{3} \pi}\right)$ $\phi=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}$ $\text { Or } \quad \phi=\frac{\pi}{6}$
AP EAMCET(Medical)-2013
Alternating Current
155024
An alternating voltage source $V=260$ sin (628t) is connected across a pure inductor of $5 \mathrm{mH}$. Inductive reactance in the circuit is:
1 $3.14 \Omega$
2 $6.28 \Omega$
3 $0.318 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{V}=260 \sin (628) \mathrm{t}$ Comparing with equation, $\mathrm{V}=\mathrm{v}_{0} \sin \omega \mathrm{t}$ $\omega=628$ $\mathrm{~L}=5 \mathrm{mH}$ We have inductive reactance, $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=628 \times 5 \times 10^{-3}=3140 \times 10^{-3}$ $\therefore \quad \mathrm{X}_{\mathrm{L}}=3.14 \Omega$
JEE Main-31.01.2023
Alternating Current
155025
What will be the self-inductance of a coil of 100 turns if a current of $5 \mathrm{~A}$ produces a magnetic flux $5 \times 10^{-5} \mathrm{~Wb}$ ?
1 $1 \mathrm{mH}$
2 $10 \mathrm{mH}$
3 $1 \mu \mathrm{H}$
4 $10 \mu \mathrm{H}$
Explanation:
A Given that, $\mathrm{N}=100$ turns Current $(\mathrm{I})=5 \mathrm{~A}$ Magnetic flux $(\phi)=5 \times 10^{-5} \mathrm{~Wb}$ We know that, Self- induction $(\mathrm{L})=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{100 \times 5 \times 10^{-5}}{5}$ $\mathrm{~L}=10^{-3} \mathrm{H}$ $\mathrm{L}=1 \mathrm{mH}$ $\left(\because 1 \mathrm{mH}=10^{-3} \mathrm{H}\right)$
MHT-CET 2011
Alternating Current
155027
If $\mathbf{N}$ is the number of turns in a coil, the value of self-inductance varies as
1 $\mathrm{N}^{\circ}$
2 $\mathrm{N}$
3 $\mathrm{N}^{2}$
4 $\mathrm{N}^{-2}$
Explanation:
C Self inductance of a solenoid is given by - $\mathrm{L}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ $\mathrm{~L} \propto \mathrm{N}^{2}$ Where, $\mathrm{N}=$ number of turns $\text { A }=\text { Area of cross-section of the solenoid }$ $l=\text { Length of the solenoid }$ $\mu_{0}=\text { Absolute permeability }$ Thus, self inductance is directly proportional to the square of the number of turns in the coil.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155023
A $0.01 \mathrm{H}$ inductor and $\sqrt{3} \pi \mathrm{ohm}$ resistance are connected in series with a $220 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}$ source. The phase difference between the current and emf is
1 $\frac{\pi}{2} \mathrm{rad}$
2 $\frac{\pi}{6} \mathrm{rad}$
3 $\frac{\pi}{3} \mathrm{rad}$
4 $\frac{\pi}{4} \mathrm{rad}$
Explanation:
B Given, Inductor $(\mathrm{L})=0.01 \mathrm{H}$ Resistance $(\mathrm{R})=\sqrt{3} \pi \Omega$ $\because$ Inductor and resistance are connected in series Then, Phase difference $(\phi)=\tan ^{-1} \frac{X_{L}}{R}$ $\phi=\tan ^{-1}\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)$ $\phi=\tan ^{-1}\left(\frac{2 \pi \mathrm{fL}}{\mathrm{R}}\right)$ Putting all the value in equation (i), we get - $\phi=\tan ^{-1}\left(\frac{2 \pi \times 50 \times 0.01}{\sqrt{3} \pi}\right)$ $\phi=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}$ $\text { Or } \quad \phi=\frac{\pi}{6}$
AP EAMCET(Medical)-2013
Alternating Current
155024
An alternating voltage source $V=260$ sin (628t) is connected across a pure inductor of $5 \mathrm{mH}$. Inductive reactance in the circuit is:
1 $3.14 \Omega$
2 $6.28 \Omega$
3 $0.318 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{V}=260 \sin (628) \mathrm{t}$ Comparing with equation, $\mathrm{V}=\mathrm{v}_{0} \sin \omega \mathrm{t}$ $\omega=628$ $\mathrm{~L}=5 \mathrm{mH}$ We have inductive reactance, $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=628 \times 5 \times 10^{-3}=3140 \times 10^{-3}$ $\therefore \quad \mathrm{X}_{\mathrm{L}}=3.14 \Omega$
JEE Main-31.01.2023
Alternating Current
155025
What will be the self-inductance of a coil of 100 turns if a current of $5 \mathrm{~A}$ produces a magnetic flux $5 \times 10^{-5} \mathrm{~Wb}$ ?
1 $1 \mathrm{mH}$
2 $10 \mathrm{mH}$
3 $1 \mu \mathrm{H}$
4 $10 \mu \mathrm{H}$
Explanation:
A Given that, $\mathrm{N}=100$ turns Current $(\mathrm{I})=5 \mathrm{~A}$ Magnetic flux $(\phi)=5 \times 10^{-5} \mathrm{~Wb}$ We know that, Self- induction $(\mathrm{L})=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{100 \times 5 \times 10^{-5}}{5}$ $\mathrm{~L}=10^{-3} \mathrm{H}$ $\mathrm{L}=1 \mathrm{mH}$ $\left(\because 1 \mathrm{mH}=10^{-3} \mathrm{H}\right)$
MHT-CET 2011
Alternating Current
155027
If $\mathbf{N}$ is the number of turns in a coil, the value of self-inductance varies as
1 $\mathrm{N}^{\circ}$
2 $\mathrm{N}$
3 $\mathrm{N}^{2}$
4 $\mathrm{N}^{-2}$
Explanation:
C Self inductance of a solenoid is given by - $\mathrm{L}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ $\mathrm{~L} \propto \mathrm{N}^{2}$ Where, $\mathrm{N}=$ number of turns $\text { A }=\text { Area of cross-section of the solenoid }$ $l=\text { Length of the solenoid }$ $\mu_{0}=\text { Absolute permeability }$ Thus, self inductance is directly proportional to the square of the number of turns in the coil.
155023
A $0.01 \mathrm{H}$ inductor and $\sqrt{3} \pi \mathrm{ohm}$ resistance are connected in series with a $220 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}$ source. The phase difference between the current and emf is
1 $\frac{\pi}{2} \mathrm{rad}$
2 $\frac{\pi}{6} \mathrm{rad}$
3 $\frac{\pi}{3} \mathrm{rad}$
4 $\frac{\pi}{4} \mathrm{rad}$
Explanation:
B Given, Inductor $(\mathrm{L})=0.01 \mathrm{H}$ Resistance $(\mathrm{R})=\sqrt{3} \pi \Omega$ $\because$ Inductor and resistance are connected in series Then, Phase difference $(\phi)=\tan ^{-1} \frac{X_{L}}{R}$ $\phi=\tan ^{-1}\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)$ $\phi=\tan ^{-1}\left(\frac{2 \pi \mathrm{fL}}{\mathrm{R}}\right)$ Putting all the value in equation (i), we get - $\phi=\tan ^{-1}\left(\frac{2 \pi \times 50 \times 0.01}{\sqrt{3} \pi}\right)$ $\phi=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}$ $\text { Or } \quad \phi=\frac{\pi}{6}$
AP EAMCET(Medical)-2013
Alternating Current
155024
An alternating voltage source $V=260$ sin (628t) is connected across a pure inductor of $5 \mathrm{mH}$. Inductive reactance in the circuit is:
1 $3.14 \Omega$
2 $6.28 \Omega$
3 $0.318 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{V}=260 \sin (628) \mathrm{t}$ Comparing with equation, $\mathrm{V}=\mathrm{v}_{0} \sin \omega \mathrm{t}$ $\omega=628$ $\mathrm{~L}=5 \mathrm{mH}$ We have inductive reactance, $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=628 \times 5 \times 10^{-3}=3140 \times 10^{-3}$ $\therefore \quad \mathrm{X}_{\mathrm{L}}=3.14 \Omega$
JEE Main-31.01.2023
Alternating Current
155025
What will be the self-inductance of a coil of 100 turns if a current of $5 \mathrm{~A}$ produces a magnetic flux $5 \times 10^{-5} \mathrm{~Wb}$ ?
1 $1 \mathrm{mH}$
2 $10 \mathrm{mH}$
3 $1 \mu \mathrm{H}$
4 $10 \mu \mathrm{H}$
Explanation:
A Given that, $\mathrm{N}=100$ turns Current $(\mathrm{I})=5 \mathrm{~A}$ Magnetic flux $(\phi)=5 \times 10^{-5} \mathrm{~Wb}$ We know that, Self- induction $(\mathrm{L})=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{100 \times 5 \times 10^{-5}}{5}$ $\mathrm{~L}=10^{-3} \mathrm{H}$ $\mathrm{L}=1 \mathrm{mH}$ $\left(\because 1 \mathrm{mH}=10^{-3} \mathrm{H}\right)$
MHT-CET 2011
Alternating Current
155027
If $\mathbf{N}$ is the number of turns in a coil, the value of self-inductance varies as
1 $\mathrm{N}^{\circ}$
2 $\mathrm{N}$
3 $\mathrm{N}^{2}$
4 $\mathrm{N}^{-2}$
Explanation:
C Self inductance of a solenoid is given by - $\mathrm{L}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ $\mathrm{~L} \propto \mathrm{N}^{2}$ Where, $\mathrm{N}=$ number of turns $\text { A }=\text { Area of cross-section of the solenoid }$ $l=\text { Length of the solenoid }$ $\mu_{0}=\text { Absolute permeability }$ Thus, self inductance is directly proportional to the square of the number of turns in the coil.
155023
A $0.01 \mathrm{H}$ inductor and $\sqrt{3} \pi \mathrm{ohm}$ resistance are connected in series with a $220 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}$ source. The phase difference between the current and emf is
1 $\frac{\pi}{2} \mathrm{rad}$
2 $\frac{\pi}{6} \mathrm{rad}$
3 $\frac{\pi}{3} \mathrm{rad}$
4 $\frac{\pi}{4} \mathrm{rad}$
Explanation:
B Given, Inductor $(\mathrm{L})=0.01 \mathrm{H}$ Resistance $(\mathrm{R})=\sqrt{3} \pi \Omega$ $\because$ Inductor and resistance are connected in series Then, Phase difference $(\phi)=\tan ^{-1} \frac{X_{L}}{R}$ $\phi=\tan ^{-1}\left(\frac{\omega \mathrm{L}}{\mathrm{R}}\right)$ $\phi=\tan ^{-1}\left(\frac{2 \pi \mathrm{fL}}{\mathrm{R}}\right)$ Putting all the value in equation (i), we get - $\phi=\tan ^{-1}\left(\frac{2 \pi \times 50 \times 0.01}{\sqrt{3} \pi}\right)$ $\phi=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}$ $\text { Or } \quad \phi=\frac{\pi}{6}$
AP EAMCET(Medical)-2013
Alternating Current
155024
An alternating voltage source $V=260$ sin (628t) is connected across a pure inductor of $5 \mathrm{mH}$. Inductive reactance in the circuit is:
1 $3.14 \Omega$
2 $6.28 \Omega$
3 $0.318 \Omega$
4 $0.5 \Omega$
Explanation:
A Given, $\mathrm{V}=260 \sin (628) \mathrm{t}$ Comparing with equation, $\mathrm{V}=\mathrm{v}_{0} \sin \omega \mathrm{t}$ $\omega=628$ $\mathrm{~L}=5 \mathrm{mH}$ We have inductive reactance, $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=628 \times 5 \times 10^{-3}=3140 \times 10^{-3}$ $\therefore \quad \mathrm{X}_{\mathrm{L}}=3.14 \Omega$
JEE Main-31.01.2023
Alternating Current
155025
What will be the self-inductance of a coil of 100 turns if a current of $5 \mathrm{~A}$ produces a magnetic flux $5 \times 10^{-5} \mathrm{~Wb}$ ?
1 $1 \mathrm{mH}$
2 $10 \mathrm{mH}$
3 $1 \mu \mathrm{H}$
4 $10 \mu \mathrm{H}$
Explanation:
A Given that, $\mathrm{N}=100$ turns Current $(\mathrm{I})=5 \mathrm{~A}$ Magnetic flux $(\phi)=5 \times 10^{-5} \mathrm{~Wb}$ We know that, Self- induction $(\mathrm{L})=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{100 \times 5 \times 10^{-5}}{5}$ $\mathrm{~L}=10^{-3} \mathrm{H}$ $\mathrm{L}=1 \mathrm{mH}$ $\left(\because 1 \mathrm{mH}=10^{-3} \mathrm{H}\right)$
MHT-CET 2011
Alternating Current
155027
If $\mathbf{N}$ is the number of turns in a coil, the value of self-inductance varies as
1 $\mathrm{N}^{\circ}$
2 $\mathrm{N}$
3 $\mathrm{N}^{2}$
4 $\mathrm{N}^{-2}$
Explanation:
C Self inductance of a solenoid is given by - $\mathrm{L}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{l}$ $\mathrm{~L} \propto \mathrm{N}^{2}$ Where, $\mathrm{N}=$ number of turns $\text { A }=\text { Area of cross-section of the solenoid }$ $l=\text { Length of the solenoid }$ $\mu_{0}=\text { Absolute permeability }$ Thus, self inductance is directly proportional to the square of the number of turns in the coil.