Answer
$ Resistance (R_{T}) $ $=$ $3.2$ $ohms$
Work Step by Step
The total power for each branch of parallel 75-Watts and 15-Watts is determined by adding these two power values.
$P_{t_{1}}$ $=$ $P_{1}$ + $P_{2}$
$P_{t_{1}}$ $=$ $75$ $Watts$ + $15$ $Watts$
$P_{t_{1}}$ $=$ $90$ $Watts$
$P_{t_{2}}$ $=$ $P_{3}$ + $P_{4}$
$P_{t_{2}}$ $=$ $75$ $Watts$ + $15$ $Watts$
$P_{t_{2}}$ $=$ $90$ $Watts$
Therefore, the total power of the circuit is:
$P_{T}$ $=$ $P_{t_{1}}$ + $P_{t_{2}}$
$P_{T}$ $=$ $90$ $Watts$ + $90$ $Watts$
$P_{T}$ $=$ $180$ $Watts$
Using the equation below: we can now determine the total resistance seen by the battery:
$ Power (P_{T}) $ $=$ $\frac{Voltage(V)^{2}}{Resistance (R_{T})}$
$ Resistance (R_{T}) $ $=$ $\frac{Voltage(V)^{2}}{Power (P_{T})}$
$ Resistance (R_{T})) $ $=$ $\frac{24^{2}}{180}$
$ Resistance (R_{T}) $ $=$ $3.2$ $ohms$
