Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watt of power. What will be the power dissipated in watt if the same resistors are connected in parallel across the same source of e.m.f. ?

Question

In a meter bridge experiment, a resistance of \( 10 \, \Omega \) is balanced by a resistance \( X \) with a balance point at \( 40 \, \mathrm{cm} \). What is the value of \( X \)?

Answer 1

To solve this problem, we need to understand how power dissipation in resistors changes when they are arranged in series versus when they are arranged in parallel. Let's tackle the problem step-by-step:

Given:

Three equal resistors, each with a resistance of \( R \), connected in series across a source of electromotive force (emf).
Total power dissipated in series configuration: 10 watts.

Step 1: Calculate the equivalent resistance when resistors are in series.

The formula for the total or equivalent resistance when resistors are connected in series is given by:

R_{\text{series}} = R + R + R = 3R

Step 2: Use the power formula to determine the emf.

Power (P) is given by the formula:

P = \frac{V^2}{R_{\text{series}}}

Given that power dissipated is 10 watts:

10 = \frac{V^2}{3R}

Solving for \( V^2 \):

V^2 = 10 \times 3R = 30R

Step 3: Calculate the equivalent resistance when resistors are in parallel.

The formula for the total resistance when resistors are connected in parallel is given by:

\frac{1}{R_{\text{parallel}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}

So, R_{\text{parallel}} = \frac{R}{3}

Step 4: Calculate the power dissipated in parallel configuration using the same emf.

Power in parallel is given by:

P_{\text{parallel}} = \frac{V^2}{R_{\text{parallel}}}

Substituting the values we have:

P_{\text{parallel}} = \frac{30R}{\frac{R}{3}} = 30R \times \frac{3}{R} = 90

Conclusion:

The power dissipated when the resistors are connected in parallel across the same source of electromotive force is 90 watts.

Answer 2