Answer
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Hint:-The electromagnetic force or e.m.f generated due to inductance will be such that the generated e.m.f will oppose the voltage difference which is present across the terminals of the inductor. As a result the flow of current will also be affected.
Formula used: The formula of the e.m.f is developed due to inductance and change in current for some interval of time is given by,
$E = - L \cdot \left( {\dfrac{{dI}}{{dt}}} \right)$
Where E is the electromagnetic force or e.m.f, L is the inductance I is the current and t is the time taken.
Complete step-by-step solution:It is given in the problem that a self-inductor generates an e.m.f of 5 volt if the change of current takes place in the time interval of 1 millisecond and the happens from 3A to 2A and we need to find out the value of self-inductance of the inductor.
As the e.m.f of the inductor is given by,
$E = - L \cdot \left( {\dfrac{{dI}}{{dt}}} \right)$
Where E is the electromagnetic force or e.m.f, L is the inductance I is the current and t is the time taken.
Let us calculate the value of the self-inductance.
$ \Rightarrow E = - L \cdot \left( {\dfrac{{dI}}{{dt}}} \right)$
Replace the value of the e.m.f equal to 5 volts $dI$ is the change in the current and $dt$ is the change in the time.
$ \Rightarrow E = - L \cdot \left( {\dfrac{{dI}}{{dt}}} \right)$
As the e.m.f is equal to $E = 5{\text{ volts}}$ the change in the current is $dI = \left( {2 - 3} \right)A$ and $dt = {10^{ - 3}}s$.
$ \Rightarrow 5 = - L \cdot \left( {\dfrac{{2 - 3}}{{{{10}^{ - 3}}}}} \right)$
$ \Rightarrow 5 = - L \cdot \left( {\dfrac{{ - 1}}{{{{10}^{ - 3}}}}} \right)$
$ \Rightarrow 5 = \dfrac{L}{{{{10}^{ - 3}}}}$
$ \Rightarrow L = 5 \times {10^{ - 3}}$
$ \Rightarrow L = 5mH$
The self-inductance is equal to$L = 5mH$. The correct answer for this problem is option D.
Note:- The negative sign in the formula for the calculation electromagnetic force or e.m.f shows that the generated e.m.f will oppose the potential difference across the terminals of the inductor. As the time increases the e.m.f generated by the inductor will increase the resistance against the potential difference.
Formula used: The formula of the e.m.f is developed due to inductance and change in current for some interval of time is given by,
$E = - L \cdot \left( {\dfrac{{dI}}{{dt}}} \right)$
Where E is the electromagnetic force or e.m.f, L is the inductance I is the current and t is the time taken.
Complete step-by-step solution:It is given in the problem that a self-inductor generates an e.m.f of 5 volt if the change of current takes place in the time interval of 1 millisecond and the happens from 3A to 2A and we need to find out the value of self-inductance of the inductor.
As the e.m.f of the inductor is given by,
$E = - L \cdot \left( {\dfrac{{dI}}{{dt}}} \right)$
Where E is the electromagnetic force or e.m.f, L is the inductance I is the current and t is the time taken.
Let us calculate the value of the self-inductance.
$ \Rightarrow E = - L \cdot \left( {\dfrac{{dI}}{{dt}}} \right)$
Replace the value of the e.m.f equal to 5 volts $dI$ is the change in the current and $dt$ is the change in the time.
$ \Rightarrow E = - L \cdot \left( {\dfrac{{dI}}{{dt}}} \right)$
As the e.m.f is equal to $E = 5{\text{ volts}}$ the change in the current is $dI = \left( {2 - 3} \right)A$ and $dt = {10^{ - 3}}s$.
$ \Rightarrow 5 = - L \cdot \left( {\dfrac{{2 - 3}}{{{{10}^{ - 3}}}}} \right)$
$ \Rightarrow 5 = - L \cdot \left( {\dfrac{{ - 1}}{{{{10}^{ - 3}}}}} \right)$
$ \Rightarrow 5 = \dfrac{L}{{{{10}^{ - 3}}}}$
$ \Rightarrow L = 5 \times {10^{ - 3}}$
$ \Rightarrow L = 5mH$
The self-inductance is equal to$L = 5mH$. The correct answer for this problem is option D.
Note:- The negative sign in the formula for the calculation electromagnetic force or e.m.f shows that the generated e.m.f will oppose the potential difference across the terminals of the inductor. As the time increases the e.m.f generated by the inductor will increase the resistance against the potential difference.
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