
When current in coil changes from $5A$ to $2A$ in $0.1s$, an average of$50V$is produced. The self-inductance of the coil is:
$(a)6H$
$(b)0.67H$
$(c)1.67H$
$(d)3H$
Answer
539.4k+ views
Hint:The general property of an inductor is that it opposes the change in current in the loop. That is, if the current in the circuit is decreasing, the inductor will convert its magnetic energy into electrical energy by creating a potential difference and if the current is increasing , then vice versa.
Complete answer:
We know, the electromotive force or E.M.F., that is the voltage drop across an inductor is given by the equation:
$\Rightarrow V=L\dfrac{di}{dt}$
$\Rightarrow V=L\dfrac{(\vartriangle i)}{(\vartriangle t)}$ [equation number$(1)$]
Where,
V = Voltage drop across the inductor
L = Self-inductance of the inductor
$di(or,\vartriangle i)$= change in current with respect to time
$dt(or,\vartriangle t)$= total time taken for change in current
Now, it’s given in the problem that:
$\Rightarrow V=-50V$ [negative sign is because voltage is produced instead of decreasing]
Also, $(\vartriangle i)$ can be calculated as follows:
$\begin{align}
& \Rightarrow (\vartriangle i)=(2-5)A \\
& \Rightarrow (\vartriangle i)=-3A \\
\end{align}$
Thus, the change in current comes out to be $-3A$ .
And, time elapsed is given by:
$\Rightarrow (\vartriangle t)=0.1s$
Putting these values of Voltage drop $(V)$ , change in current $(\vartriangle i)$ and time elapsed $(t)$ in equation (1), we get:
\[\Rightarrow -50=L\left( \dfrac{-3}{0.1} \right)\]
On cross multiplying and rearranging terms, we get:
$\Rightarrow L=\dfrac{5}{3}H$
$\Rightarrow L=1.67H$
Hence, the self-inductance of the inductor comes out to be $1.67H$.
Hence, option $(c)$ is the correct option.
Note:
We should always keep in mind the general properties of equipment which are commonly used in circuit problems like, resistor, capacitor and inductor. Their common properties can be like, their characteristics under different parameters like pressure and temperature. Are they extensive or intensive properties, etc.
Complete answer:
We know, the electromotive force or E.M.F., that is the voltage drop across an inductor is given by the equation:
$\Rightarrow V=L\dfrac{di}{dt}$
$\Rightarrow V=L\dfrac{(\vartriangle i)}{(\vartriangle t)}$ [equation number$(1)$]
Where,
V = Voltage drop across the inductor
L = Self-inductance of the inductor
$di(or,\vartriangle i)$= change in current with respect to time
$dt(or,\vartriangle t)$= total time taken for change in current
Now, it’s given in the problem that:
$\Rightarrow V=-50V$ [negative sign is because voltage is produced instead of decreasing]
Also, $(\vartriangle i)$ can be calculated as follows:
$\begin{align}
& \Rightarrow (\vartriangle i)=(2-5)A \\
& \Rightarrow (\vartriangle i)=-3A \\
\end{align}$
Thus, the change in current comes out to be $-3A$ .
And, time elapsed is given by:
$\Rightarrow (\vartriangle t)=0.1s$
Putting these values of Voltage drop $(V)$ , change in current $(\vartriangle i)$ and time elapsed $(t)$ in equation (1), we get:
\[\Rightarrow -50=L\left( \dfrac{-3}{0.1} \right)\]
On cross multiplying and rearranging terms, we get:
$\Rightarrow L=\dfrac{5}{3}H$
$\Rightarrow L=1.67H$
Hence, the self-inductance of the inductor comes out to be $1.67H$.
Hence, option $(c)$ is the correct option.
Note:
We should always keep in mind the general properties of equipment which are commonly used in circuit problems like, resistor, capacitor and inductor. Their common properties can be like, their characteristics under different parameters like pressure and temperature. Are they extensive or intensive properties, etc.
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