Question

An indicator of 20 H and a resistance of $10\Omega$, are connected to a bettery of 5 volt in series, then initial rate of change of current is :-

• A. $1.5amp/s$
• B. $2.0amp/s$
• C. $2.5amp/s$
• D. $0.25amp/s$

Answer 1

The correct option is D $0.25amp/s$

Given,

$L=20H$

$V=5V$

$R=10\Omega$

The current flowing in the $L-R$ circuit is,

$I=I_0(1-e^{-Lt/R})$

$\frac{dI}{dt}=\frac{I_{0}R}{L}{e}^{Lt/R}$

The initial rate of change of current is

$(\frac{dI}{dt}{)}_0=\frac{I_{0}R}{L}=\frac{V}{L}$

$(\frac{dI}{dt}{)}_0=\frac{5}{20}=0.25amp/s$

The correct option is D.