358379
An emf of \(15\;\,V\) is applied in a circuit containing \(5\,H\) inductance and \(10 \Omega\), the ratio of currents at time \(t=\infty\) and \(t = 1\;\,s\) is
The curent through the circuit is \(i=i_{o}\left(1-e^{\frac{-R}{L} t}\right)=i_{o}\left(1-e^{-2 t}\right)\) At \(t=1 s i_{1}=i_{o}\left(1-\dfrac{1}{e^{2}}\right)\) As \(t \rightarrow \infty\) \(\dfrac{i_{2}}{i_{1}}=\dfrac{e^{2}}{e^{2}-1} \Rightarrow \phi=L_{1} i_{1}=L_{2} i_{2}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358380
In the branch \({A B}\) of a circuit, as shown in the figure, a current \({I=(t+2) A}\) is flowing where \({t}\) is the time in second. At \({t=0}\), the value of \({\left(V_{A}-V_{B}\right)}\) will be
1 \({3 V}\)
2 \({-3 V}\)
3 \({-5 V}\)
4 \({5 V}\)
Explanation:
On applying \({K V L}\) rule, \({V_{A}-I R-L \dfrac{d I}{d t}+10=V_{B} \Rightarrow V_{A}-V_{B}=-3 V}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358381
A coil of inductance \(8.4mH\) and resistance \(6 \Omega\) is connected to a \(12 \mathrm{~V}\) battery. The current in the coil is \(1.0\;A\) in the time (apporox.)
1 \(20\sec \)
2 35 milli sec
3 \(500\sec \)
4 1 milli sec
Explanation:
The current through the coil is \(\begin{aligned}& i=\dfrac{\varepsilon}{R}\left[1-e^{\frac{-R}{L} t}\right]=\dfrac{12}{6}\left[1-e^{\frac{-R}{L} t}\right] \\& 1=2\left[1-e^{\frac{-R}{L} t}\right] \\& e^{\frac{-R}{L} t}=\dfrac{1}{2} \Rightarrow t=\dfrac{\ln (2) L}{R}=1 m s\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358382
The current in a \(L R\) circuit builds up to \(\frac{3}{4}th\) of its steady state value in \(4\;s\). The time constant of this circuit is
1 \(\dfrac{4}{\operatorname{In} 2} s\)
2 \(\dfrac{3}{\operatorname{In} 3} s\)
3 \(\dfrac{2}{\operatorname{In} 2} s\)
4 \(\dfrac{1}{\operatorname{In} 2} s\)
Explanation:
We know that \(i=i_{0}\left[1-e^{\frac{-R t}{L}}\right]\) \(\text { or } \dfrac{3}{4} i_{0}=i_{0}\left[1-e^{-t / \tau}\right]\) (Where \(\tau=\dfrac{L}{R}=\) time constant ) \(\begin{aligned}& \dfrac{3}{4}=1-e^{-t / \tau} \text { or }^{-t / \tau}=\dfrac{1}{4} \\& e^{t / \tau}=4 \text { or } \dfrac{t}{\tau}=\operatorname{In} 4 \\& \Rightarrow \tau=\dfrac{t}{\operatorname{In} 4}=\dfrac{4}{2 \operatorname{In} 2} \Rightarrow \tau=\dfrac{2}{\operatorname{In} 2} \mathrm{sec} .\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358383
In the circuit given below, switch \(S\) is closed for sufficiently long time. At \(t=0\), the switch is opened, then
1 The time constant is \(L /\left(R_{1}+R_{2}\right)\)
2 The current after one time constant is \(E / e\left(R_{1}+R_{2}\right)\)
3 The time constant is \(\dfrac{L\left(R_{1} R_{2}\right)}{R_{1}+R_{2}}\)
4 The current after one time constant is \(2 E / e\left(R_{1}+R_{2}\right)\)
Explanation:
Current through the inductor at \(t = 0,\) \(i_{0}=\dfrac{E}{R_{1}}\) \(\therefore\) Current at time \(t\) \(\begin{aligned}& i=i_{0} e^{-t / \tau} \\& i=i_{o} e^{-1}=\dfrac{E}{e R_{1}} \quad(\text { At } t=\tau)\end{aligned}\) Also \(\quad \tau=\dfrac{L}{R_{1}+R_{2}}\)
358379
An emf of \(15\;\,V\) is applied in a circuit containing \(5\,H\) inductance and \(10 \Omega\), the ratio of currents at time \(t=\infty\) and \(t = 1\;\,s\) is
The curent through the circuit is \(i=i_{o}\left(1-e^{\frac{-R}{L} t}\right)=i_{o}\left(1-e^{-2 t}\right)\) At \(t=1 s i_{1}=i_{o}\left(1-\dfrac{1}{e^{2}}\right)\) As \(t \rightarrow \infty\) \(\dfrac{i_{2}}{i_{1}}=\dfrac{e^{2}}{e^{2}-1} \Rightarrow \phi=L_{1} i_{1}=L_{2} i_{2}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358380
In the branch \({A B}\) of a circuit, as shown in the figure, a current \({I=(t+2) A}\) is flowing where \({t}\) is the time in second. At \({t=0}\), the value of \({\left(V_{A}-V_{B}\right)}\) will be
1 \({3 V}\)
2 \({-3 V}\)
3 \({-5 V}\)
4 \({5 V}\)
Explanation:
On applying \({K V L}\) rule, \({V_{A}-I R-L \dfrac{d I}{d t}+10=V_{B} \Rightarrow V_{A}-V_{B}=-3 V}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358381
A coil of inductance \(8.4mH\) and resistance \(6 \Omega\) is connected to a \(12 \mathrm{~V}\) battery. The current in the coil is \(1.0\;A\) in the time (apporox.)
1 \(20\sec \)
2 35 milli sec
3 \(500\sec \)
4 1 milli sec
Explanation:
The current through the coil is \(\begin{aligned}& i=\dfrac{\varepsilon}{R}\left[1-e^{\frac{-R}{L} t}\right]=\dfrac{12}{6}\left[1-e^{\frac{-R}{L} t}\right] \\& 1=2\left[1-e^{\frac{-R}{L} t}\right] \\& e^{\frac{-R}{L} t}=\dfrac{1}{2} \Rightarrow t=\dfrac{\ln (2) L}{R}=1 m s\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358382
The current in a \(L R\) circuit builds up to \(\frac{3}{4}th\) of its steady state value in \(4\;s\). The time constant of this circuit is
1 \(\dfrac{4}{\operatorname{In} 2} s\)
2 \(\dfrac{3}{\operatorname{In} 3} s\)
3 \(\dfrac{2}{\operatorname{In} 2} s\)
4 \(\dfrac{1}{\operatorname{In} 2} s\)
Explanation:
We know that \(i=i_{0}\left[1-e^{\frac{-R t}{L}}\right]\) \(\text { or } \dfrac{3}{4} i_{0}=i_{0}\left[1-e^{-t / \tau}\right]\) (Where \(\tau=\dfrac{L}{R}=\) time constant ) \(\begin{aligned}& \dfrac{3}{4}=1-e^{-t / \tau} \text { or }^{-t / \tau}=\dfrac{1}{4} \\& e^{t / \tau}=4 \text { or } \dfrac{t}{\tau}=\operatorname{In} 4 \\& \Rightarrow \tau=\dfrac{t}{\operatorname{In} 4}=\dfrac{4}{2 \operatorname{In} 2} \Rightarrow \tau=\dfrac{2}{\operatorname{In} 2} \mathrm{sec} .\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358383
In the circuit given below, switch \(S\) is closed for sufficiently long time. At \(t=0\), the switch is opened, then
1 The time constant is \(L /\left(R_{1}+R_{2}\right)\)
2 The current after one time constant is \(E / e\left(R_{1}+R_{2}\right)\)
3 The time constant is \(\dfrac{L\left(R_{1} R_{2}\right)}{R_{1}+R_{2}}\)
4 The current after one time constant is \(2 E / e\left(R_{1}+R_{2}\right)\)
Explanation:
Current through the inductor at \(t = 0,\) \(i_{0}=\dfrac{E}{R_{1}}\) \(\therefore\) Current at time \(t\) \(\begin{aligned}& i=i_{0} e^{-t / \tau} \\& i=i_{o} e^{-1}=\dfrac{E}{e R_{1}} \quad(\text { At } t=\tau)\end{aligned}\) Also \(\quad \tau=\dfrac{L}{R_{1}+R_{2}}\)
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PHXII06:ELECTROMAGNETIC INDUCTION
358379
An emf of \(15\;\,V\) is applied in a circuit containing \(5\,H\) inductance and \(10 \Omega\), the ratio of currents at time \(t=\infty\) and \(t = 1\;\,s\) is
The curent through the circuit is \(i=i_{o}\left(1-e^{\frac{-R}{L} t}\right)=i_{o}\left(1-e^{-2 t}\right)\) At \(t=1 s i_{1}=i_{o}\left(1-\dfrac{1}{e^{2}}\right)\) As \(t \rightarrow \infty\) \(\dfrac{i_{2}}{i_{1}}=\dfrac{e^{2}}{e^{2}-1} \Rightarrow \phi=L_{1} i_{1}=L_{2} i_{2}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358380
In the branch \({A B}\) of a circuit, as shown in the figure, a current \({I=(t+2) A}\) is flowing where \({t}\) is the time in second. At \({t=0}\), the value of \({\left(V_{A}-V_{B}\right)}\) will be
1 \({3 V}\)
2 \({-3 V}\)
3 \({-5 V}\)
4 \({5 V}\)
Explanation:
On applying \({K V L}\) rule, \({V_{A}-I R-L \dfrac{d I}{d t}+10=V_{B} \Rightarrow V_{A}-V_{B}=-3 V}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358381
A coil of inductance \(8.4mH\) and resistance \(6 \Omega\) is connected to a \(12 \mathrm{~V}\) battery. The current in the coil is \(1.0\;A\) in the time (apporox.)
1 \(20\sec \)
2 35 milli sec
3 \(500\sec \)
4 1 milli sec
Explanation:
The current through the coil is \(\begin{aligned}& i=\dfrac{\varepsilon}{R}\left[1-e^{\frac{-R}{L} t}\right]=\dfrac{12}{6}\left[1-e^{\frac{-R}{L} t}\right] \\& 1=2\left[1-e^{\frac{-R}{L} t}\right] \\& e^{\frac{-R}{L} t}=\dfrac{1}{2} \Rightarrow t=\dfrac{\ln (2) L}{R}=1 m s\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358382
The current in a \(L R\) circuit builds up to \(\frac{3}{4}th\) of its steady state value in \(4\;s\). The time constant of this circuit is
1 \(\dfrac{4}{\operatorname{In} 2} s\)
2 \(\dfrac{3}{\operatorname{In} 3} s\)
3 \(\dfrac{2}{\operatorname{In} 2} s\)
4 \(\dfrac{1}{\operatorname{In} 2} s\)
Explanation:
We know that \(i=i_{0}\left[1-e^{\frac{-R t}{L}}\right]\) \(\text { or } \dfrac{3}{4} i_{0}=i_{0}\left[1-e^{-t / \tau}\right]\) (Where \(\tau=\dfrac{L}{R}=\) time constant ) \(\begin{aligned}& \dfrac{3}{4}=1-e^{-t / \tau} \text { or }^{-t / \tau}=\dfrac{1}{4} \\& e^{t / \tau}=4 \text { or } \dfrac{t}{\tau}=\operatorname{In} 4 \\& \Rightarrow \tau=\dfrac{t}{\operatorname{In} 4}=\dfrac{4}{2 \operatorname{In} 2} \Rightarrow \tau=\dfrac{2}{\operatorname{In} 2} \mathrm{sec} .\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358383
In the circuit given below, switch \(S\) is closed for sufficiently long time. At \(t=0\), the switch is opened, then
1 The time constant is \(L /\left(R_{1}+R_{2}\right)\)
2 The current after one time constant is \(E / e\left(R_{1}+R_{2}\right)\)
3 The time constant is \(\dfrac{L\left(R_{1} R_{2}\right)}{R_{1}+R_{2}}\)
4 The current after one time constant is \(2 E / e\left(R_{1}+R_{2}\right)\)
Explanation:
Current through the inductor at \(t = 0,\) \(i_{0}=\dfrac{E}{R_{1}}\) \(\therefore\) Current at time \(t\) \(\begin{aligned}& i=i_{0} e^{-t / \tau} \\& i=i_{o} e^{-1}=\dfrac{E}{e R_{1}} \quad(\text { At } t=\tau)\end{aligned}\) Also \(\quad \tau=\dfrac{L}{R_{1}+R_{2}}\)
358379
An emf of \(15\;\,V\) is applied in a circuit containing \(5\,H\) inductance and \(10 \Omega\), the ratio of currents at time \(t=\infty\) and \(t = 1\;\,s\) is
The curent through the circuit is \(i=i_{o}\left(1-e^{\frac{-R}{L} t}\right)=i_{o}\left(1-e^{-2 t}\right)\) At \(t=1 s i_{1}=i_{o}\left(1-\dfrac{1}{e^{2}}\right)\) As \(t \rightarrow \infty\) \(\dfrac{i_{2}}{i_{1}}=\dfrac{e^{2}}{e^{2}-1} \Rightarrow \phi=L_{1} i_{1}=L_{2} i_{2}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358380
In the branch \({A B}\) of a circuit, as shown in the figure, a current \({I=(t+2) A}\) is flowing where \({t}\) is the time in second. At \({t=0}\), the value of \({\left(V_{A}-V_{B}\right)}\) will be
1 \({3 V}\)
2 \({-3 V}\)
3 \({-5 V}\)
4 \({5 V}\)
Explanation:
On applying \({K V L}\) rule, \({V_{A}-I R-L \dfrac{d I}{d t}+10=V_{B} \Rightarrow V_{A}-V_{B}=-3 V}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358381
A coil of inductance \(8.4mH\) and resistance \(6 \Omega\) is connected to a \(12 \mathrm{~V}\) battery. The current in the coil is \(1.0\;A\) in the time (apporox.)
1 \(20\sec \)
2 35 milli sec
3 \(500\sec \)
4 1 milli sec
Explanation:
The current through the coil is \(\begin{aligned}& i=\dfrac{\varepsilon}{R}\left[1-e^{\frac{-R}{L} t}\right]=\dfrac{12}{6}\left[1-e^{\frac{-R}{L} t}\right] \\& 1=2\left[1-e^{\frac{-R}{L} t}\right] \\& e^{\frac{-R}{L} t}=\dfrac{1}{2} \Rightarrow t=\dfrac{\ln (2) L}{R}=1 m s\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358382
The current in a \(L R\) circuit builds up to \(\frac{3}{4}th\) of its steady state value in \(4\;s\). The time constant of this circuit is
1 \(\dfrac{4}{\operatorname{In} 2} s\)
2 \(\dfrac{3}{\operatorname{In} 3} s\)
3 \(\dfrac{2}{\operatorname{In} 2} s\)
4 \(\dfrac{1}{\operatorname{In} 2} s\)
Explanation:
We know that \(i=i_{0}\left[1-e^{\frac{-R t}{L}}\right]\) \(\text { or } \dfrac{3}{4} i_{0}=i_{0}\left[1-e^{-t / \tau}\right]\) (Where \(\tau=\dfrac{L}{R}=\) time constant ) \(\begin{aligned}& \dfrac{3}{4}=1-e^{-t / \tau} \text { or }^{-t / \tau}=\dfrac{1}{4} \\& e^{t / \tau}=4 \text { or } \dfrac{t}{\tau}=\operatorname{In} 4 \\& \Rightarrow \tau=\dfrac{t}{\operatorname{In} 4}=\dfrac{4}{2 \operatorname{In} 2} \Rightarrow \tau=\dfrac{2}{\operatorname{In} 2} \mathrm{sec} .\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358383
In the circuit given below, switch \(S\) is closed for sufficiently long time. At \(t=0\), the switch is opened, then
1 The time constant is \(L /\left(R_{1}+R_{2}\right)\)
2 The current after one time constant is \(E / e\left(R_{1}+R_{2}\right)\)
3 The time constant is \(\dfrac{L\left(R_{1} R_{2}\right)}{R_{1}+R_{2}}\)
4 The current after one time constant is \(2 E / e\left(R_{1}+R_{2}\right)\)
Explanation:
Current through the inductor at \(t = 0,\) \(i_{0}=\dfrac{E}{R_{1}}\) \(\therefore\) Current at time \(t\) \(\begin{aligned}& i=i_{0} e^{-t / \tau} \\& i=i_{o} e^{-1}=\dfrac{E}{e R_{1}} \quad(\text { At } t=\tau)\end{aligned}\) Also \(\quad \tau=\dfrac{L}{R_{1}+R_{2}}\)
358379
An emf of \(15\;\,V\) is applied in a circuit containing \(5\,H\) inductance and \(10 \Omega\), the ratio of currents at time \(t=\infty\) and \(t = 1\;\,s\) is
The curent through the circuit is \(i=i_{o}\left(1-e^{\frac{-R}{L} t}\right)=i_{o}\left(1-e^{-2 t}\right)\) At \(t=1 s i_{1}=i_{o}\left(1-\dfrac{1}{e^{2}}\right)\) As \(t \rightarrow \infty\) \(\dfrac{i_{2}}{i_{1}}=\dfrac{e^{2}}{e^{2}-1} \Rightarrow \phi=L_{1} i_{1}=L_{2} i_{2}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358380
In the branch \({A B}\) of a circuit, as shown in the figure, a current \({I=(t+2) A}\) is flowing where \({t}\) is the time in second. At \({t=0}\), the value of \({\left(V_{A}-V_{B}\right)}\) will be
1 \({3 V}\)
2 \({-3 V}\)
3 \({-5 V}\)
4 \({5 V}\)
Explanation:
On applying \({K V L}\) rule, \({V_{A}-I R-L \dfrac{d I}{d t}+10=V_{B} \Rightarrow V_{A}-V_{B}=-3 V}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358381
A coil of inductance \(8.4mH\) and resistance \(6 \Omega\) is connected to a \(12 \mathrm{~V}\) battery. The current in the coil is \(1.0\;A\) in the time (apporox.)
1 \(20\sec \)
2 35 milli sec
3 \(500\sec \)
4 1 milli sec
Explanation:
The current through the coil is \(\begin{aligned}& i=\dfrac{\varepsilon}{R}\left[1-e^{\frac{-R}{L} t}\right]=\dfrac{12}{6}\left[1-e^{\frac{-R}{L} t}\right] \\& 1=2\left[1-e^{\frac{-R}{L} t}\right] \\& e^{\frac{-R}{L} t}=\dfrac{1}{2} \Rightarrow t=\dfrac{\ln (2) L}{R}=1 m s\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358382
The current in a \(L R\) circuit builds up to \(\frac{3}{4}th\) of its steady state value in \(4\;s\). The time constant of this circuit is
1 \(\dfrac{4}{\operatorname{In} 2} s\)
2 \(\dfrac{3}{\operatorname{In} 3} s\)
3 \(\dfrac{2}{\operatorname{In} 2} s\)
4 \(\dfrac{1}{\operatorname{In} 2} s\)
Explanation:
We know that \(i=i_{0}\left[1-e^{\frac{-R t}{L}}\right]\) \(\text { or } \dfrac{3}{4} i_{0}=i_{0}\left[1-e^{-t / \tau}\right]\) (Where \(\tau=\dfrac{L}{R}=\) time constant ) \(\begin{aligned}& \dfrac{3}{4}=1-e^{-t / \tau} \text { or }^{-t / \tau}=\dfrac{1}{4} \\& e^{t / \tau}=4 \text { or } \dfrac{t}{\tau}=\operatorname{In} 4 \\& \Rightarrow \tau=\dfrac{t}{\operatorname{In} 4}=\dfrac{4}{2 \operatorname{In} 2} \Rightarrow \tau=\dfrac{2}{\operatorname{In} 2} \mathrm{sec} .\end{aligned}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358383
In the circuit given below, switch \(S\) is closed for sufficiently long time. At \(t=0\), the switch is opened, then
1 The time constant is \(L /\left(R_{1}+R_{2}\right)\)
2 The current after one time constant is \(E / e\left(R_{1}+R_{2}\right)\)
3 The time constant is \(\dfrac{L\left(R_{1} R_{2}\right)}{R_{1}+R_{2}}\)
4 The current after one time constant is \(2 E / e\left(R_{1}+R_{2}\right)\)
Explanation:
Current through the inductor at \(t = 0,\) \(i_{0}=\dfrac{E}{R_{1}}\) \(\therefore\) Current at time \(t\) \(\begin{aligned}& i=i_{0} e^{-t / \tau} \\& i=i_{o} e^{-1}=\dfrac{E}{e R_{1}} \quad(\text { At } t=\tau)\end{aligned}\) Also \(\quad \tau=\dfrac{L}{R_{1}+R_{2}}\)
