Question

Illustration 36.
Three identical bulbs are connected in a circuit as shown. What will happen to the
nnected in a circuit as shown. What will happen to the brightness of bulbs A & B if
the switch is closed ?
​

Answer 1

Answer:

Hii Mate,

The brightness of bulbs A & B is Greater than Bulb C..

Answer 2

Concept

Our perception of brightness does not depend only on the intensity of the light but also on the wavelength of the light. So, two lights of the same intensity but different wavelengths have different perceived brightness.

The light bulb emits radiation in the visible region also. When the temperature of the filament increases the intensity of lights of all wavelengths increases. So, if the temperature of the filament goes up we perceive the bulb as more bright.

The temperature of the bulb is a function of the power delivered to it by the electricity. So, in order to look at the brightness of bulbs A and B, we need to consider their power.

The electric bulbs have resistance due to which they dissipate electrical energy. So, their power is given by the following relation

                                 P = VI = V²/R = I² R.

Given

• A, B, and C are identical bulbs.

• Bulbs A, B, and C are in series.

• Voltage V is applied across a series combination of bulbs.

Find

The variation in brightness of bulbs A and B after the switch S is closed.

Solution

Voltage across the bulbs when S is open

A, B, and C are identical so their resistances are the same.

The voltage V is, therefore, distributed equally across all the bulbs. So, the voltage across each bulb is V/3.

Power dissipated by bulbs when S is open

The Power dissipated by each bulb, P = (V/3)² /R = V²/(9R).         ...(1)

Voltage across bulbs A and B when S is closed

When S is closed, the two ends of the bulb C will be at the same potential so potential difference or the voltage across C is zero.

Only A and B will share the voltage V. Here also the voltage across A and B are the same and are equal to V/2.

Power dissipated by A and B when S is closed

Power dissipated by each of A and B is P’ = (V/2)²/R = V²/(4R).   ...(2)

Relation between the two powers

From (1) and (2),

     9P = 4P'

⇒ P' = 9P/4.

Relation between their brightness

Assuming a direct relation between the brightness and power dissipation, the brightness of the bulbs A and B is increased by a factor of 9/4.

Brightness of the bulbs A and B are increased by a factor of 9/4.

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