Question

compared to a what is the brightness of each bulb in b

compared to a what is the brightness of each bulb in c

In the figure below, all light bulbs are identical and behave like resistors. All batteries are identical. Ideal wires connect the light bulbs to the batteries in three different ways: (a), (b), (c). 00,0 00.0 000009 000 (a) (b) (c)

Answer 1

The energy supplied to the bulb by the resistor is converted into light. The intensity of the light of the bulb is proportional to the power (energy per second) delivered to the bulb. Let R be the resistance of a bulb, and V be the voltage of the battery. In the first case, the battery is connected across a single bulb, the power delivered is

$P_a=VI=V\times\frac VR=\frac{V^2}{R}$

2 Pai R

In the second circuit, two bulbs are connected in series, their equivalent resistance is

$R_{eq}=2R$

The series combination of the two bulbs is connected across the battery, the power delivered by the two bulbs is

$P_{2b}=\frac{V^2}{R_{eq}}=\frac{V^2}{2R}$

The two bulbs are identical, the power delivered to each bulb is equal. The power delivered to single bulb is half of the power delivered to both bulbs.

$P_b=\frac{P_{2b}}{2}=\frac{V^2}{4R}$

Using (1)

$P_b=\frac{P_a}{4}$

Since intensity is proportional to the power delivered, the brightness of a single bulb in (b) is one-forth of the brightness of the bulb in (a).

In circuit (c), the two bulbs are in parallel combination with the battery. The voltage across each bulb is V. So, the power delivered to each bulb in circuit (c) is

$P_c=\frac{V^2}{R}$

Using (1)

$P_c=P_a$

Since intensity is proportional to the power delivered, the brightness of a single bulb in (c) is equal to the brightness of the bulb in (a).