Two identical resistors, each of resistance 15 W, are connected in (i) series, and (ii) parallel, in turn to a battery of 6 V. Calculate the ratio of the power consumed in the combination of resistors in each case.
Given:
Resistance, $R_1=R_2=15\Omega$
Voltage, $V=6V$.
To find: The ratio of power consumed in the combination of resistors in each case, $\frac {P_S}{P_R}$.
Solution: (i) Series case
In series, total resistance is given as-
$R_S=R_1+R_2$
$R_S=15+15$
$R_S=30\Omega$
Thus, the total resistance in series, $R_S$ is $30\Omega$
Now,
We know that, power consumed by a resistor is given as-
$Power=\frac{Voltage^2}{Resistance}$ Or, $P = \frac{V^2}{R}$
Substituting the value of $V$ and $R$ we get-
$P_S=\frac{6^2}{30}$
$P_S=\frac{36}{30}$
$P_S=1.2W$
Thus, the power consumed in series, $P_S$ is 1.2 watt.
(ii) Parallel case:
In parallel, total resistance is given as-
$\frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}$
$\frac{1}{R_P}=\frac{1}{15}+\frac{1}{15}$
$\frac{1}{R_P}=\frac{1+1}{15}$
$\frac{1}{R_P}=\frac{2}{15}$
$R_P=\frac{15}{2}$
$R_P=7.5\Omega$
Thus, the effective resistance in parallel, $R_P$ is $7.5\Omega$
Now,
We know that, power consumed by a resistor is given as-
$Power=\frac{Voltage^2}{Resistance}$ Or, $P = \frac{V^2}{R}$
Substituting the value of $V$ and $R$ we get-
$P_R=\frac{6^2}{7.5}$
$P_R=\frac{36}{7.5}$
$P_R=\frac{360}{75}$
$P_R=4.8W$
Thus, the power consumed in parallel, $P_R$ is 4.8 watt.
Ratio of power consumed in the two combinations,
$\frac {P_S}{P_R}=\frac{1.2\ W}{4.8 \ W}=\frac{1}{4}=1:4$
Thus, the ratio of power consumed in the combination of resistors in each case is 1:4.
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