Question

Two identical resistors each of resistance $2\Omega$ are connected in turn a. in series and b. in parallel to a battery of 12 V. Calculate the ratio of power consumed in the two cases.

Answer 1

Step 1:Given data

Resistance of each resistor $R=2\Omega$

Number of resistors = 2

Voltage of battery $V=12V$

Step 2: Finding effective resistance

Case a:

Let $R_{s_{}}$be the effective resistance of the series combination of resistors.

Effective resistance in the case of the series combination of resistors

$$\begin{gathered} R_s=R+R \\ {R}_s=2R \\ {R}_s=2\times 2\Omega \\ {R}_s=4\Omega \end{gathered}$$

Case b:

Let $R_p$ be the effective resistance of the parallel combination of resistors.

In the case of parallel combination, the reciprocal of effective resistance

$$\begin{gathered} \frac{1}{R_p}=\frac{1}{R}+\frac{1}{R} \\ \frac{1}{R_p}=\frac{2}{R} \\ \frac{1}{R_p}=\frac{2}{2\Omega } \\ \frac{1}{R_p}=\frac{1}{1\Omega } \end{gathered}$$

Effective resistance in the case of the parallel combination $R_p=1\Omega$

Step 3: Equations for the power consumed in each case

In both series and parallel combinations, the battery remains the same. Hence, the value of V is the same.

Power consumed in a series combination of resistors = $P_s=\frac{V^2}{R_s}$

Power consumed in a parallel combination of resistors = $P_p=\frac{V^2}{R_p}$

Step 4: Finding the ratio of power consumed in each case

The ratio of power consumed in two cases = $\frac{P_s}{P_p}=\frac{{\displaystyle \frac{V^2}{R_s}}}{\frac{V^2}{R_p}}=\frac{V^2}{R_s}\times \frac{R_p}{V^2}=\frac{R_p}{R_s}=\frac{1\Omega }{4\Omega }=\frac{1}{4}$

$P_s:P_p=1:4$

Final answer: The ratio of power consumed in series combination to that in parallel combination is 1: 4.