358550
If the inductance of a coil is 1 henry, its effective resistance in a \(DC\) circuit will be
1 \({\infty}\)
2 zero
3 \(1\,\Omega \)
4 \(2\,\Omega \)
Explanation:
Given inductance \({L=1 {H}} \) Its inductive reactance \({=X_{L}=\omega L=2 \pi v L}\) For \({D C v=0}\) \({\therefore \quad X_{L}=0 \rightarrow}\) effective resistance in \({D C}\) circuit. So correct option is (2)
PHXII06:ELECTROMAGNETIC INDUCTION
358551
A coil of \(N = 100\) turns carries a current \(I = 5\;A\) and creates a magnetic flux \(\phi = {10^{ - 5}}T{m^{ - 2}}\)per turn. The value of its inductance \(L\) will be
1 \(0.15\,mH\)
2 \(0.05\,mH\)
3 \(0.10\,mH\)
4 \(0.20\,mH\)
Explanation:
\(N\phi = Li \Rightarrow 100 \times {10^{ - 5}} = L \times 5 \Rightarrow L = 0.2mH\)
PHXII06:ELECTROMAGNETIC INDUCTION
358552
If in a coil rate of change of area is \(5\;{m^2}/s\) millisecond and current becomes \(1\,amp\) from 2 \(amp\) in \(2 \times {10^{ - 3}}\sec \). If magnitude of field is 1 tesla then self-inductance of the coil is
1 \(5\,H\)
2 \(2\,H\)
3 \(10\,H\)
4 \(20\,H\)
Explanation:
\(N \phi=L i \Rightarrow \dfrac{N d \phi}{d t}=\dfrac{L d i}{d t} \Rightarrow N B \dfrac{d A}{d t}=\dfrac{L d i}{d t}\) \(\Rightarrow \dfrac{1 \times 1 \times 5}{10^{-3}}=L \times\left(\dfrac{2-1}{2 \times 10^{-3}}\right) \Rightarrow L=10 H\)
PHXII06:ELECTROMAGNETIC INDUCTION
358553
A coil of \(N = 100\) turns carries a current \(I = 5\;A\) and creates a magnetic flux \(\phi = {10^{ - 5}}T{m^{ - 2}}\) per turn. The value of its inductance \(L\) will be:
1 \(0.05\,mH\)
2 \(0.10\,mH\)
3 \(0.15\,mH\)
4 \(0.2\,mH\)
Explanation:
\(N \phi=L i\) \( \Rightarrow 100 \times {10^{ - 5}} = L \times 5 \Rightarrow L = 0.2mH\)
358550
If the inductance of a coil is 1 henry, its effective resistance in a \(DC\) circuit will be
1 \({\infty}\)
2 zero
3 \(1\,\Omega \)
4 \(2\,\Omega \)
Explanation:
Given inductance \({L=1 {H}} \) Its inductive reactance \({=X_{L}=\omega L=2 \pi v L}\) For \({D C v=0}\) \({\therefore \quad X_{L}=0 \rightarrow}\) effective resistance in \({D C}\) circuit. So correct option is (2)
PHXII06:ELECTROMAGNETIC INDUCTION
358551
A coil of \(N = 100\) turns carries a current \(I = 5\;A\) and creates a magnetic flux \(\phi = {10^{ - 5}}T{m^{ - 2}}\)per turn. The value of its inductance \(L\) will be
1 \(0.15\,mH\)
2 \(0.05\,mH\)
3 \(0.10\,mH\)
4 \(0.20\,mH\)
Explanation:
\(N\phi = Li \Rightarrow 100 \times {10^{ - 5}} = L \times 5 \Rightarrow L = 0.2mH\)
PHXII06:ELECTROMAGNETIC INDUCTION
358552
If in a coil rate of change of area is \(5\;{m^2}/s\) millisecond and current becomes \(1\,amp\) from 2 \(amp\) in \(2 \times {10^{ - 3}}\sec \). If magnitude of field is 1 tesla then self-inductance of the coil is
1 \(5\,H\)
2 \(2\,H\)
3 \(10\,H\)
4 \(20\,H\)
Explanation:
\(N \phi=L i \Rightarrow \dfrac{N d \phi}{d t}=\dfrac{L d i}{d t} \Rightarrow N B \dfrac{d A}{d t}=\dfrac{L d i}{d t}\) \(\Rightarrow \dfrac{1 \times 1 \times 5}{10^{-3}}=L \times\left(\dfrac{2-1}{2 \times 10^{-3}}\right) \Rightarrow L=10 H\)
PHXII06:ELECTROMAGNETIC INDUCTION
358553
A coil of \(N = 100\) turns carries a current \(I = 5\;A\) and creates a magnetic flux \(\phi = {10^{ - 5}}T{m^{ - 2}}\) per turn. The value of its inductance \(L\) will be:
1 \(0.05\,mH\)
2 \(0.10\,mH\)
3 \(0.15\,mH\)
4 \(0.2\,mH\)
Explanation:
\(N \phi=L i\) \( \Rightarrow 100 \times {10^{ - 5}} = L \times 5 \Rightarrow L = 0.2mH\)
358550
If the inductance of a coil is 1 henry, its effective resistance in a \(DC\) circuit will be
1 \({\infty}\)
2 zero
3 \(1\,\Omega \)
4 \(2\,\Omega \)
Explanation:
Given inductance \({L=1 {H}} \) Its inductive reactance \({=X_{L}=\omega L=2 \pi v L}\) For \({D C v=0}\) \({\therefore \quad X_{L}=0 \rightarrow}\) effective resistance in \({D C}\) circuit. So correct option is (2)
PHXII06:ELECTROMAGNETIC INDUCTION
358551
A coil of \(N = 100\) turns carries a current \(I = 5\;A\) and creates a magnetic flux \(\phi = {10^{ - 5}}T{m^{ - 2}}\)per turn. The value of its inductance \(L\) will be
1 \(0.15\,mH\)
2 \(0.05\,mH\)
3 \(0.10\,mH\)
4 \(0.20\,mH\)
Explanation:
\(N\phi = Li \Rightarrow 100 \times {10^{ - 5}} = L \times 5 \Rightarrow L = 0.2mH\)
PHXII06:ELECTROMAGNETIC INDUCTION
358552
If in a coil rate of change of area is \(5\;{m^2}/s\) millisecond and current becomes \(1\,amp\) from 2 \(amp\) in \(2 \times {10^{ - 3}}\sec \). If magnitude of field is 1 tesla then self-inductance of the coil is
1 \(5\,H\)
2 \(2\,H\)
3 \(10\,H\)
4 \(20\,H\)
Explanation:
\(N \phi=L i \Rightarrow \dfrac{N d \phi}{d t}=\dfrac{L d i}{d t} \Rightarrow N B \dfrac{d A}{d t}=\dfrac{L d i}{d t}\) \(\Rightarrow \dfrac{1 \times 1 \times 5}{10^{-3}}=L \times\left(\dfrac{2-1}{2 \times 10^{-3}}\right) \Rightarrow L=10 H\)
PHXII06:ELECTROMAGNETIC INDUCTION
358553
A coil of \(N = 100\) turns carries a current \(I = 5\;A\) and creates a magnetic flux \(\phi = {10^{ - 5}}T{m^{ - 2}}\) per turn. The value of its inductance \(L\) will be:
1 \(0.05\,mH\)
2 \(0.10\,mH\)
3 \(0.15\,mH\)
4 \(0.2\,mH\)
Explanation:
\(N \phi=L i\) \( \Rightarrow 100 \times {10^{ - 5}} = L \times 5 \Rightarrow L = 0.2mH\)
358550
If the inductance of a coil is 1 henry, its effective resistance in a \(DC\) circuit will be
1 \({\infty}\)
2 zero
3 \(1\,\Omega \)
4 \(2\,\Omega \)
Explanation:
Given inductance \({L=1 {H}} \) Its inductive reactance \({=X_{L}=\omega L=2 \pi v L}\) For \({D C v=0}\) \({\therefore \quad X_{L}=0 \rightarrow}\) effective resistance in \({D C}\) circuit. So correct option is (2)
PHXII06:ELECTROMAGNETIC INDUCTION
358551
A coil of \(N = 100\) turns carries a current \(I = 5\;A\) and creates a magnetic flux \(\phi = {10^{ - 5}}T{m^{ - 2}}\)per turn. The value of its inductance \(L\) will be
1 \(0.15\,mH\)
2 \(0.05\,mH\)
3 \(0.10\,mH\)
4 \(0.20\,mH\)
Explanation:
\(N\phi = Li \Rightarrow 100 \times {10^{ - 5}} = L \times 5 \Rightarrow L = 0.2mH\)
PHXII06:ELECTROMAGNETIC INDUCTION
358552
If in a coil rate of change of area is \(5\;{m^2}/s\) millisecond and current becomes \(1\,amp\) from 2 \(amp\) in \(2 \times {10^{ - 3}}\sec \). If magnitude of field is 1 tesla then self-inductance of the coil is
1 \(5\,H\)
2 \(2\,H\)
3 \(10\,H\)
4 \(20\,H\)
Explanation:
\(N \phi=L i \Rightarrow \dfrac{N d \phi}{d t}=\dfrac{L d i}{d t} \Rightarrow N B \dfrac{d A}{d t}=\dfrac{L d i}{d t}\) \(\Rightarrow \dfrac{1 \times 1 \times 5}{10^{-3}}=L \times\left(\dfrac{2-1}{2 \times 10^{-3}}\right) \Rightarrow L=10 H\)
PHXII06:ELECTROMAGNETIC INDUCTION
358553
A coil of \(N = 100\) turns carries a current \(I = 5\;A\) and creates a magnetic flux \(\phi = {10^{ - 5}}T{m^{ - 2}}\) per turn. The value of its inductance \(L\) will be:
1 \(0.05\,mH\)
2 \(0.10\,mH\)
3 \(0.15\,mH\)
4 \(0.2\,mH\)
Explanation:
\(N \phi=L i\) \( \Rightarrow 100 \times {10^{ - 5}} = L \times 5 \Rightarrow L = 0.2mH\)
