The self-induced emf of a coil is 25,V. When | vedclass

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(A) The self-induced $emf$ is given by $\varepsilon = L \frac{di}{dt}$.Given $\varepsilon = 25\,V$,$\Delta i = 25\,A - 10\,A = 15\,A$,and $\Delta t =

Vedclass · Accepted Answer

$437.5$

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(A) The self-induced $emf$ is given by $\varepsilon = L \frac{di}{dt}$.Given $\varepsilon = 25\,V$,$\Delta i = 25\,A - 10\,A = 15\,A$,and $\Delta t = 1\,s$.Thus,$L = \frac{\varepsilon \Delta t}{\Delta i} = \frac{25 	imes 1}{15} = \frac{5}{3}\,H$.The change in energy stored in the inductor is $\Delta U = \frac{1}{2} L (i_f^2 - i_i^2)$.Substituting the values: $\Delta U = \frac{1}{2} 	imes \frac{5}{3} 	imes (25^2 - 10^2)$.$\Delta U = \frac{5}{6} 	imes (625 - 100) = \frac{5}{6} 	imes 525$.$\Delta U = 437.5\,J$.

The self-induced $emf$ of a coil is $25\,V$. When the current in it is changed at a uniform rate from $10\,A$ to $25\,A$ in $1\,s$,the change in the energy stored in the inductor is......$J$.

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