If two bulbs, one of 200 W and the other of 100 W are connected in series with a 100 V battery, then which bulb will have more brightness:

CONCEPT:

Electric power:

• The rate at which the electric energy is dissipated or consumed is termed as electric power.
• The electric power is given as,

\(⇒ P=IV=I^{2}R=\frac{V^{2}}{R}\)

Where P = electric power, V = voltage, I = current and R = resistance

CALCULATION:

Given P₁ = 200 W, P₂ = 100 W and V = 100 V,

We know that for an electric bulb the resistance is given as,

\(⇒ R=\frac{V^{2}}{P}\)

\(⇒ R\propto\frac{1}{P}\)     -----(1)

• For an electric bulb, the resistance of the bulb is inversely proportional to the power of the bulb.
• So the bulb which has more power will have low resistance, therefore the 100 W bulb will have more resistance compared to the 200 W bulb.
• The heat dissipated by the bulb is given as,

⇒ H = I²R     -----(2)

• Both the bulbs are connected in series so the current in both the bulbs will be equal. So the heat dissipated will be more in the bulb which has more resistance.
• Since the resistance of the 100 W bulb is more, so the heat dissipation of the 100 W bulb will be more.
• The brightness of the bulb depends on the heat dissipation by the bulb, so the 100 W bulb will have more brightness.