In the circuit shown below, the current through the $12$ resistor is $0\mathrm{.}24$ A $.$

$($a$)$ Determine the terminal voltage of the source, and

$($b$)$ the power dissipated in the $15$ resistor.

$($c$)$ Assume the emf of the battery to be $45$ V $,$ and calculate the internal resistance of the battery.

Solution:

$($a$)V=40\mathrm{.}32V$

$($b$)P=4\mathrm{.}98W$

$($c$)r=1\mathrm{.}625$