Question

Two identical resistors, each of resistance $15\mathrm{ohm}$, are connected in (i) series, and (ii) parallel, in turn to a battery of $6\mathrm{V}$. Calculate the ratio of the power consumed in the combination of resistors in each case.

Answer 1

Step 1: Calculation of equivalent resistance while connected in series

The equivalent resistance of the resistors when connected in series can be found as follows:

$$\begin{gathered} R=R_1+R_2 \\ R=15\mathrm{ohm}+15\mathrm{ohm} \\ R=30\mathrm{ohm} \end{gathered}$$

Where $R$ is the equivalent resistance, $R_1$ and $R_2$ is the resistance of the resistors.

Step 2: Calculation of equivalent resistance while connected in parallel

The equivalent resistance of the resistors when connected in parallel can be found as follows:

$$\begin{gathered} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R}=\frac{1}{15\mathrm{ohm}}+\frac{1}{15\mathrm{ohm}} \\ \frac{1}{R}=\frac{2}{15\mathrm{ohm}} \\ R=7.5\mathrm{ohm} \end{gathered}$$

Step 3: Calculation of power in both the cases

The equation of power can be expressed as follows:

$P=\frac{V^2}{R}$

Where, $P$ is the power, and $V$ is the voltage.

1. Power while the connection is in series

$$\begin{gathered} P_{\mathrm{series}}=\frac{{\left(6\mathrm{V}\right)}^2}{30\mathrm{ohm}} \\ {P}_{\mathrm{series}}=1.2\mathrm{watt} \end{gathered}$$

2. Power while the connection is in parallel

$$\begin{gathered} P_{\mathrm{parallel}}=\frac{{\left(6\mathrm{V}\right)}^2}{7.5\mathrm{ohm}} \\ {P}_{\mathrm{parallel}}=4.8\mathrm{watt} \end{gathered}$$

3. Calculation of ratio of powers

$$\begin{gathered} \frac{P_{\mathrm{series}}}{P_{\mathrm{parallel}}}=\frac{1.2}{4.8} \\ \frac{P_{\mathrm{series}}}{P_{\mathrm{parallel}}}=\frac{1}{4} \end{gathered}$$

Therefore, the ratio of powers is $1:4$.