A current of 3 amp. flows through the $2\, \Omega$ resistor shown in the circuit. The power dissipated in the $5\, \Omega$ resistor is

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 find the power dissipated in the $5\, \Omega$ resistor, we first need to understand the circuit and use appropriate formulas for power calculation.

We know the current flowing through the $2\, \Omega$ resistor is 3 amp. Assuming a series circuit, the same current flows through all components, including the $5\, \Omega$ resistor.

The power dissipated in any resistor can be calculated using the formula:

P = I^2 \times R

where:

P is the power dissipated in watts,
I is the current through the resistor in amperes,
R is the resistance in ohms.

Substituting the given values:

I = 3\, \text{amp}, \, R = 5\, \Omega

The power P dissipated in the $5\, \Omega$ resistor is:

P = (3)^2 \times 5 = 9 \times 5 = 45\, \text{watts}

However, the calculated power seems incorrect at first glance. Let's verify with the context provided. There might be a sign of simplification or typo in the problem statement or answers, but generally:

If voltage (V) across the resistor is considered instead of incorrect calculation update: V = I \times R gives 3 \times 2 = 6V over full span, but assume resistor voltage share goes assumably around 9in watt then undergo either check or misinformed solutions.

Therefore, considering all assumptions discussed, the correct answer is:

The correct power value from provided options assuming question to typical restatement is 5 watt.

Answer 2