Question

In the circuit shown, the key(K) is closed at t=0, the current through the key at the $t={10}^{-1}$ ln 2s; is.

• A. 2A
• B. 3.5A
• C. 2.5A
• D. zero

Answer 1

Answer: (A)

2A

In $LR$ circuit current is,

$i=i_0{e}^{-\frac{t}{\tau }}$

Time constant is

$\tau =\frac{L}{R_{eq}}$

$R_{eq}=\frac{10\times 10}{10+10}=5\Omega$

$\tau =\frac{5\times {10}^{-3}}{5}={10}^{-3}$

$i=\frac{V}{R}=\frac{20}{6+4}=2A$