Chapter 23.2, Problem 29SSC

To determine

To compare: the brightness of the bulbs shown in the figure below.

Explanation of Solution

Given:

Four resistors are connected.

The circuit diagram is

  

Introduction:

The brightness of a bulb is proportional to its power dissipation. Since it is given that the bulbs are identical, the resistance of the bulb will be the same.

The current $I_3$ that reaches point C, have two paths to follow. The amount of current flows through a branch depends on the equivalent resistance of that branch.

The current $I$ is given by $I=\frac{V}{R}$ .

Here, $V$ and $R$ are potential difference and resistance.

That is, the current is inversely proportional to resistance.

On comparing the two branches of the parallel circuit, we can find that the equivalent resistance of branch having bulb 1 will be less than that of the branch having bulb 2 and 3. Thus, $I_1$ will be greater than $I_2$ .

We have a formula for power that is, $P=I^{2}R$ .

As the bulbs are identical, power depends only on $I$ . Thus, among the three bulbs, the current through the bulb 1 is higher and hence it will be the brightest one.

Since the bulbs 2 and 3 are connected in series, the same current flows through them. Thus, bulbs 2 and 3 will have the same brightness.

Conclusion:

Hence, bulbs 2 and 3 will have the same brightness. Bulb 1 will be less than that of the branch having bulb 2 and 3.