Question

An inductor L = 50 mH carrying an initial current $I_0=2.5$ A is connected across a non linear resistor with p.d varying with current I according to $V=10I^2$. How long after (in ms) current though inductor becomes 1.25A

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

Given $-L\frac{dI}{dt}=10l^2$
$-{\int }_{2.5}^{1.25}L\frac{dI}{I^2}=10{\int }_0^{t}dt$
$\Rightarrow t=2ms$