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# 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.**