Question

An ideal coil of $10$ Henry is joined in series with a resistance of $5$ ohm and a battery of $5$ volt. $2$ second after joining, the current flowing in ampere in the circuit will be :

• A. $e^{-1}$
• B. $1-e^{-1}$
• C. $1-e$
• D. $e$

Answer 1

Answer: (B)

$1-e^{-1}$

$I=I_0(1-e^{t/\tau })I_0=\frac{E}{R}=\frac{5}{5}=1amp$

$\tau =\frac{L}{R}=\left(\frac{10}{5}\right)=2secI=(1-e^{-2/2})=(1-e^{-1})$