# Which Bulb Glows Brighter When Connected in Series and Parallel & Why?

**Two Bulbs of 80W & 100W are Connected in Series & Parallel – Which One will Glow Brighter?**

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 brighter**.

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Table of Contents

**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.

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**Power**

**P = V x I** or **P = I ^{2 }R** or

**P = V**

^{2}/RNow, 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} = V_{1} + V_{2} + V_{3} …+ V_{n}

Current in series circuit: I_{T} = I_{1} = I_{2} = I_{3} …I_{n}

Therefore_{, }

R = V^{2} / P_{80}

R_{80W }= 220^{2} / 80W

**R _{80W }= 605Ω**

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

R = V^{2} / P_{100}

R_{100W }= 220^{2} / 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 = I^{2}R

P_{80W }= (0.202A)^{2} x 605Ω

**P _{80W }= 24.68 W**

Power dissipated by Bulb 2 (100W)

P = I^{2}R_{100}

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** … (Basic Ohm’s Law)

For Bulb 1 (80W)

V_{80} = I x R_{80} = 0.202 x 605Ω = 122.3V

**V _{80} = 122.3V**

For Bulb 2 (100W)

V_{100} = I x R_{100} = 0.202 x 484Ω = 97.7V

**V _{100} = 97.7V**

Now,

Power dissipated by Bulb 1 (80W)

P = V^{2}_{80}/R_{80}

P_{80W }= 122.3^{2}V / 605Ω

**P _{80W }= 24.7 W**

Power dissipated by Bulb 2 (100W)

P = V^{2}_{100}/R_{100}

P_{100W }= 97.72^{2}V / 484Ω

**P _{100W }= 19.74 W**

Total Voltage in the series circuit

V_{T} = V_{80} + V_{100} = 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.

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**When Bulbs are Connected in Parallel**

**Ratings of bulbs Wattage are different and connected in the parallel circuit:**

Now we have the same two bulbs each of **80W** (Bulb 1) and **100W** (Bulb 2) **connected in parallel** across the supply voltage of 220V AC. In that case, the same will happen i.e. **the bulb with more current and high power dissipation will glow brighter** than the other one. This time, **100W Bulb (2) will glow brighter and bulb 1 of 80W will dimmer**. In short, In parallel, both bulbs have the same voltage across them. The bulb with the lower resistance will conduct more current and therefore have a higher power dissipation and brightness. Confused? as the case has been reversed. Let see the below calculations and examples to clear the confusion.

Power

**P = V x I** or **P = I ^{2 }R** or

**P = V**

^{2}/RNow, **the resistance of Bulb 1 (80W)**;

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We know that voltages are the same in the parallel circuit and the rated voltage of bulbs are 220V. i.e.

Voltage in Parallel Circuit: V_{T} = V_{1} = V_{2} = V3 …V_{n}

Current in parallel circuit: I_{T} = I_{1} + I_{2} + I_{3} …I_{n}

Therefore_{, }

R = V^{2} / P

R_{80W }= 220^{2} / 80W

**R _{80W }= 605Ω**

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

R = V^{2} / P

R_{100W }= 220^{2} / 100W

**R _{100W }= 484Ω**

Now,

Power dissipated by Bulb 1 (80W) as voltages are same in a parallel circuit.

P = V^{2}/R_{1}

P_{80W }= (220V)^{2} / 605Ω

**P _{80W} = 80 W**

Power dissipated by Bulb 2 (100W)

P = V^{2}/R_{2}

P_{100W }= (220V)^{2} / 484Ω

**P _{100W }= 100 W**

Hence, proved **P _{100W }> P**

_{80W }i.e.

**Bulb 2 (100W) is greater in power dissipation than bulb 1 (80W)**. Therefore, the

**100W bulb is brighter than 80W bulb when connected in parallel.**

To verify the above case, You may also find the current for each bulb and then find the power dissipation by **P = V x I** as follows. We used the rated voltage of the bulb which is 220V.

I = P / V

For Bulb 1 (80W)

I_{80} = P_{80} / 220 = 80W / 220 = 0.364A

**I _{80} = 0.364A**

For Bulb 2 (100W)

I_{100} = P_{100} / 220 = 100W / 220 = 0.455A

**I _{100} = 0.455A**

Now,

Power dissipated by Bulb 1 (80W) as voltages are same in the parallel circuit.

P = I^{2}R_{1}

P_{80W }= 0.364^{2}A x 605Ω

P_{80W} = 80 W

Power dissipated by Bulb 2 (100W)

P = I^{2}R_{2}

P_{100W }= 0.455^{2}A x 484Ω

**P _{100W} = 100 W**

Total Current in the parallel circuit

I_{T} = I_{1} + I_{2} = 0.364 + 0.455 = 0.818A

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

**Without Calculations & Examples**

Calculations and examples are for newbies. To make it simple, keep the fact in mind that always, **The bulb with a “high power” will have “less resistance”**. **The filament of the bulb with a high rating is thicker than the lower wattage**. In our case, the filament of the 80W bulb is thinner than the 100W bulb.

In other words, **100 Watt bulb has less resistance and 80 Watt bulb has a high resistance**.

**When bulbs connected in Series**

We know that current in a series circuit is same at each point mean both bulbs getting the same current and voltages are different. Obliviously, the voltage drop across higher resistance bulb (80W) will be more. So the **80W bulb will glow brighter as compared to 100W bulb connected in series** because the same current is flowing through both of the bulbs where the 80W bulb has more resistance due to lower wattage as the filament is thinner means it dissipates more power (**P=V ^{2}/R where power is directly proportional to the voltage and inversely proportional to the resistance**) and

**produce higher heat**

**& light**than the 100 W bulb.

**When Bulbs are connected in Parallel**

We also know that voltage in a parallel circuit is the same at each section which means both of the bulbs have the same voltage drop. Now more current will flow in the bulb which has less resistance which is 100W bulb this time which means 100W bulb dissipate more power than 80W bulb (**P=I ^{2}R**) where current and resistance are directly proportional to the power. Hence,

**100W bulb will glow brighter in a parallel**

**circuit**.

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- Difference Between Series and Parallel Circuit – Comparison

**How to know if Bulbs are Connected in series or Parallel?**

Most of the household electrical wiring & installation are wired in parallel or series-parallel instead of series as parallel wiring has some advantages over a series wiring. So we may notice that higher rated bulb glows more brightly as compared to lower wattage rated bulbs. In that case, 100W bulb glows more brightly than 60W or 80W bulb.

Now, You should know that the light bulb with higher power rating will glow brighter when connected in parallel and the light bulb with less power rating will glow brighter in case of series wiring and Vice versa.

**Key Points**:

- In a series circuit, 80W bulb glows brighter due to high power dissipation instead of a 100W bulb.
- In a parallel circuit, 100W bulb glows brighter due to high power dissipation instead of an 80W bulb.
- The bulb which dissipates more power will glow brighter.
- 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.
- In parallel, both bulbs have the same voltage across them. The bulb with the lower resistance will conduct more current and therefore have a higher power dissipation and brightness.
- Most household electrical wiring bulbs are wired in parallel.

**Note & Good to know:**

- Temperature changes in real life bulbs, so Ohm’s law is not applicable as it is applicable when resistance is constant where resistance depends on temperature.
- The temperature coefficient of the bulbs should be taken into the account. We neglect the temperature coefficient to use the Ohm’s law for simplification.
- In case of incandescent lamps and tungsten lamp filament, incandescent is a non-linear device (resistance) which has a positive temperature coefficient.

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- How To wire Lamps in Series?
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But what’s about leds. Is this eqally applicable for leds ?

Pls tell me.

LEDs are current-operated devices (diodes) and as such only drop a certain voltage across them (.7 – 1 volt typically for small diodes.) They need an external current-limiting device such as a resistor or they will immediately go into thermal runaway and burn out.

thanks alot for your support

REALLY helped. Great job!!

Voltage in series circuit: VT = V1 = V2 = V3 …Vn

* Voltage in series circuit: VT = V1 +V2 +V3 +…Vn

just correct this error in typing in above article

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: VT = V1 = V2 = V3 …Vn

The above expression is wrong

is should be : VT = V1+ V2 + V3 …Vn

As mentioned, Voltage are additive in series circuit, so it is a typo which is corrected now. Any way, Thanks for the attention.

please reply

why are both 80w and 100w bubs in series connection having voltage of 220

arent devices connected in series required to split the total voltage , then how can it be said that both bulbs have 220V

thanks a lot. it was really helpful

This is not a base-level question, so think beyond the explanation above. And by the way I don’t have a good answer.

re: lightbulbs in series. A bulb’s resistance increases proportional to temperature, so as a bulb heats up, its increasing resistance acts as an automatic current limiter. With multiple bulbs in series, why does not the highest-resistance bulb not get hotter and hotter until it fails, with the other bulbs getting colder and colder as the first bulb’s resistance keeps going up and reducing current to them?

I am thinking that as the filament gets hotter, it radiates more heat, limiting its temperature to an equilibrium point. Ditto for the lower-resistance bulbs, whose equilibrium point is at a lower temperature. Thoughts?