Question

Two bulbs of 40 W and 60 W are connected in series to an external potential difference. Which bulb will glow brighter? Why?

Answer 1

Let the external potential difference be 230 V

For 40 W bulb resistance is R

For 60 W bulb resistance is R

According to Ohm’s law

I = \(\frac{V}{R}\)

Current flowing through 40 W bulb is

\(\frac{230}{1322.5}\) = 0.1739 A

Current flowing through 60 W bulb is

\(\frac{230}{881.6}\) = 0.2608 A

When bulbs are connected in series effective resistance is 

R_S = R₁ + R₂ = 1322.5 + 881.6 

R_S = 2204.1Ω

Net current

I = \(\frac{230}{2204.1}\) = 0.1043 A

Using power equation P = I²R 

For 40 W bulb P = I²R 

= (0.1043)² × 1322.5

= (0.01087) × 1322.5 = 14.386 W 

For 60 W bulb P = I²R 

= (0.1043)² × 881.6 

= (0.01087) × 881.6 = 9.5904 W 

In a series system, higher the resistance, higher the brightness so, 40 W bulb glows brighter.