Answer
(a) The equivalent resistance is $23~\Omega$
(b) $I = 8~A$
Work Step by Step
(a) We can find the equivalent resistance of the $12~\Omega$ and the $24~\Omega$ resistors:
$\frac{1}{R'} = \frac{1}{12~\Omega}+ \frac{1}{24~\Omega}$
$\frac{1}{R'} = \frac{1}{8~\Omega}$
$R' = 8~\Omega$
We can find the equivalent resistance with the $15~\Omega$ resistor:
$R_{eq} = 8~\Omega+ 15~\Omega$
$R_{eq} = 23~\Omega$
The equivalent resistance is $23~\Omega$
(b) We can find the total current in the circuit:
$I = \frac{V}{R_{eq}}$
$I = \frac{276~V}{23~\Omega}$
$I = 12~A$
We can find the potential difference across the $15~\Omega$ resistor:
$\Delta V = (12~A)(15~\Omega)$
$\Delta V = 180~V$
Thus the potential difference across the $12~\Omega$ resistor is $276~V-180~V$ which is $96~V$. We can find the current in the $12~\Omega$ resistor:
$I = \frac{\Delta V}{R}$
$I = \frac{96~V}{12~\Omega}$
$I = 8~A$
