Answer
(a) $4.0\times 10^{-4}s $
(b) $57mA $
(c) $65mA $
Work Step by Step
(a) We know that $ R^{\prime}=R_1+R_2$
The equivalent resistance of the branch circuit with resistors $ R^{\prime}$, $ R_3$ in parallel is given as
$\frac{1}{R^{\prime \prime}}=\frac{1}{R_3}+\frac{1}{R^{\prime}}$
$\frac{1}{R^{\prime \prime}}=\frac{1}{R_3}+\frac{1}{R_1+R_2}$
As $ R_1=R_2=R_3=R $
$\frac{1}{R^{\prime \prime}}=\frac{1}{R}+\frac{1}{R+R}=\frac{3}{2}R $
$\implies R^{\prime \prime}=\frac{2}{3}R $
Now the equivalent resistance of the two resistors, $ R $, $ R^{\prime \prime}$ in series is given as
$ R_{eq}=R+R^{\prime \prime}=R+\frac{2}{3}R=\frac{5R}{3}$
$\implies R_{eq}=\frac{5(55\Omega)}{3}=91.67\Omega $
Now $ t=\frac{L}{R_{eq}}$
We plug in the known values to obtain:
$ t=\frac{37\times 10^{-3}H}{91.67\Omega}=4.0\times 10^{-4}s $
(b) We know that
$ I=\frac{\epsilon}{R_{eq}}(1-e^{-2})$
We plug in the known values to obtain:
$ I=(\frac{6.0V}{91.67\Omega})(1-0.1353)$
$ I=57mA $
(c) We know that
$ I=\frac{\epsilon}{R_{eq}}$
We plug in the known values to obtain:
$ I=\frac{6.0V}{91.67\Omega}$
$ I=65mA $
