ELECTRICAL TECHNOLOGY

The most confusing question we received that if two bulbs are connected in series and then in parallel, which one will glow brighter and what are the exact reasons? Well, there are lots of info around the web, but we will go in a very step by step details to calculate the exact values to clear the confusion.

First of all, keep in mind that the bulb having a high resistance and dissipate more power in the circuit (no matter series or parallel) will glow brighter. In other words, the brightness of the bulb depends on voltage, current (V x I = Power) as well as resistance.

Also, keep in mind that power dissipated in Watts is not the unit of brightness. Unit of brightness is lumens (denoted by lm which is SI derived unit of luminous flux) also known as candela (base unit of luminous intensity). But the light brightness is directly proportional to the bulb wattage. That’s why the more wattage a bulb is using will glow brighte

When Bulbs are Connected in Series

Ratings of bulbs Wattage are different and connected in a series circuit:

Suppose we have two bulbs each of 80W (Bulb 1) and 100W (Bulb 2), rated voltages of both bulbs are 220V and connected in series with a supply voltage of 220V AC. In that case, the bulb with high resistance and more power dissipation will glow brighter than the other one. i.e. 80W Bulb (1) will glow brighter and bulb (2) of 100W will dimmer in series connection. In short, In series, both bulbs have the same current flowing through them. The bulb with the higher resistance will have a greater voltage drop across it and therefore have a higher power dissipation and brightness.How? Let see the below calculations and examples.

Power

P = V x I or P = I² R or P = V²/R

Now, the resistance of Bulb 1 (80W);

We know that current is same and voltage are additive in a series circuit but the rated voltage of bulbs are 220V. i.e.

Voltage in series circuit: V_T = V₁ = V₂ = V₃ …V_n

Current in series circuit: I_T = I₁ = I₂ = I₃ …I_n

Therefore_,

R = V² / P₈₀

R_80W = 220² / 80W

R_80W = 605Ω

And, the resistance of Bulb 2 (100W);

R = V² / P₁₀₀

R_100W = 220² / 100W

R_100W = 484Ω

Now, Current;

I = V/R

= V / (R_80W + R_100W)

= 220V / (605Ω + 484Ω)

I = 0.202A

Now,

Power dissipated by Bulb 1 (80W)

P = I²R

P_80W = (0.202A)² x 605Ω

P_80W = 24.68 W

Power dissipated by Bulb 2 (100W)

P = I²R₁₀₀

P_100W = (0.202A)² x 484Ω

P_100W = 19.74 W

Hence, proved power dissipated P_80W > P_100W i.e. Bulb 1 (80W) is greater in power dissipation than bulb 2 (100W). Therefore, the 80W bulb is brighter than 100W bulb when connected in series.

You may also find the voltage drop across each bulb and then find the power dissipation by P = V x I as follows to verify the case

V = I x R or I = V/R or R = V/I

For Bulb 1 (80W)

V₈₀ = I x R₈₀ = 0.202 x 605Ω = 122.3V

V₈₀ = 122.3V

For Bulb 2 (100W)

V₁₀₀ = I x R₁₀₀ = 0.202 x 484Ω = 97.7V

V₁₀₀ = 97.7V

Now,

Power dissipated by Bulb 1 (80W)

P = V²₈₀/R₈₀

P_80W = 122.3²V / 605Ω

P_80W = 24.7 W

Power dissipated by Bulb 2 (100W)

P = V²₁₀₀/R₁₀₀

P_100W = 97.72²V / 484Ω

P_100W = 19.74 W

Total Voltage in the series circuit

V_T = V₈₀ + V₁₀₀ = 122.3 + 97.7 = 220V

Again proved that 80W bulb is greater in power dissipation than 100W bulb when connected in series. Hence, 80W bulb will glow brighter than 100W bulb when connected in series