Answer
$R_{1}$ $=$ $ 7.68 $ $ ohms $
$R_{2}$ $=$ $ 5.76 $ $ ohms $
$R_{T}$ $=$ $ 3.29 $ $ohms$
Work Step by Step
According to the problem, the load should be two 75-W headlight, but one 75-W and one 100-W was placed instead. To determine the resistance of each load, we should look at the following formulas:
$Equation$ $1$:
$ Power (P) $ $=$ $Voltage (V)$ $\times$ $ Current(I )$
$Equation$ $2$:
$ Current(I )$ $=$ $\frac{Voltage (V)}{Resistance (R)}$
$Equation$ $3$ (substituting formula of $ Current (I) $ from $Equation$ $2$):
$ Power (P) $ $=$ $Voltage (V)$ $\times$ $\frac{Voltage (V)}{Resistance (R)}$
Simplifying $Equation$ $3$:
$ Power (P) $ $=$ $\frac{Voltage(V)^{2}}{Resistance (R)}$
$Equation$ $4$: (manipulating $Equation$ $3$ for the formula of $ Resistance (R) $)
$ Resistance (R) $ $=$ $\frac{Voltage(V)^{2}}{Power (P)}$
With given values from the problem:
$P_{1}$ $=$ $ 75 $ $ Watts $
$P_{2}$ $=$ $ 100 $ $ Watts$
$ V $ $=$ $ 24 $ $ Volts $
Substituting the given to $Equation$ $4$:
$ Resistance $ $(R_{1})$ $=$ $\frac{24^{2}}{75}$
$R_{1}$ $=$ $ 7.68 $ $ ohms $
$ Resistance $ $(R_{2})$ $=$ $\frac{24^{2}}{100}$
$R_{2}$ $=$ $ 5.76 $ $ ohms $
To find the total resistance seen by the battery, we will compute the total resistance of 2 parallel resistors by using the formula below:
$Equation$ $5$:
$ Resistance $ $(R_{T})$ $=$ $\frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$
$R_{T}$ $=$ $\frac{7.68 \times 5.76}{7.68 + 5.76}$
$R_{T}$ $=$ $ 3.29 $ $ohms$
