KVPY Sample Paper KVPY Stream-SX Model Paper-28

The self produced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is:

A) \[740\text{ }J\]

B) \[437.5\text{ }J\]

C) \[540\text{ }J\]

D) \[637.5\text{ }J\]

Correct Answer: B

Solution :

\[\frac{L\Delta T}{\Delta t}=25\]

\[\Rightarrow \] \[L=\frac{25\times 1}{15}=\frac{5}{3}\]

\[\Delta U=\frac{1}{2}L(I_{f}^{2}-I_{1}^{2})=\frac{1}{2}\times \frac{5}{3}\times ({{25}^{2}}-{{10}^{2}})\]

\[=\frac{5}{6}\times 525=437.5J.\]