Answer
(a) $I_C\lt I_B=I_A$
(b) $t_B\lt t_A\lt t_C$
Work Step by Step
(a) We know that
$I_A=\frac{\epsilon_A}{R_A}$
$\implies I_A=\frac{12V}{4\Omega}=3.0A$
The initial current in circuit B is
$I_B=\frac{\epsilon_B}{R_B}$
$I_B=\frac{9V}{3\Omega}$
$I_B=3.0A$
The initial current in circuit C is
$I_C=\frac{9V}{9}=1.0A$
Thus, the increasing order of the circuits based on their initial currents is
$I_C\lt I_B=I_A$.
(b) We know that
$t=-RC\ln(0.5)$
For circuit A
$t_A=-R_AC_A\ln (0.5)$
$t_A=-(4\Omega)(3\mu F)\ln(0.5)$
$t_A=8.32\mu s$
For circuit B
$t_B=-R_BC_B\ln(0.5)$
$\implies t_B=-(3\Omega)(1\mu F)\ln(0.5)$
$t_B=2.08\mu s$
For circuit C
$t_C=-R_C C_c\ln(0.5)$
$\implies t_C=-(9\Omega)(2\mu F)\ln(0.5)$
$t_C=12.48\mu s$
Thus, the required increasing order is
$t_B\lt t_A\lt t_C$
