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 = 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: VT = V1 + V2 + V3 …+ Vn
Current in series circuit: IT = I1 = I2 = I3 …In
Therefore,
R = V2 / P80
R80W = 2202 / 80W
R80W = 605Ω
And, the resistance of Bulb 2 (100W);
R = V2 / P100
R100W = 2202 / 100W
R100W = 484Ω
Now, Current;
I = V/R
= V / (R80W + R100W)
= 220V / (605Ω + 484Ω)
I = 0.202A
Now,
Power dissipated by Bulb 1 (80W)
P = I2R
P80W = (0.202A)2 x 605Ω
P80W = 24.68 W
Power dissipated by Bulb 2 (100W)
P = I2R100
P100W = (0.202A)2 x 484Ω
P100W = 19.74 W
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Hence, proved power dissipated P80W > P100W 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)
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
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Now,
Power dissipated by Bulb 1 (80W)
P = V280/R80
P80W = 122.32V /605Ω
P80W = 24.7 W
Power dissipated by Bulb 2 (100W)
P = V2100/R100
P100W = 97.722V / 484Ω
P100W = 19.74 W
Total Voltage in the series circuit
VT = 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.
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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 = I2 R or P = V2/R
Now, 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: VT = V1 = V2 = V3 …Vn
Current in parallel circuit: IT = I1 + I2 + I3 …In
Therefore,
R = V2 / P
R80W = 2202 / 80W
R80W = 605Ω
And, the resistance of Bulb 2 (100W);
R = V2 / P
R100W = 2202 / 100W
R100W = 484Ω
Now,
Power dissipated by Bulb 1 (80W) as voltages are same in a parallel circuit.
P = V2/R1
P80W = (220V)2 / 605Ω
P80W = 80 W
Power dissipated by Bulb 2 (100W)
P = V2/R2
P100W = (220V)2 / 484Ω
P100W = 100 W
Hence, proved P100W > P80W 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)
I80 = P80 / 220 = 80W / 220 = 0.364A
I80 = 0.364A
For Bulb 2 (100W)
I100 = P100 / 220 = 100W / 220 = 0.455A
I100 = 0.455A
Now,
Power dissipated by Bulb 1 (80W) as voltages are same in the parallel circuit.
P = I2R1
P80W = 0.3642A x 605Ω
P80W = 80 W
Power dissipated by Bulb 2 (100W)
P = I2R2
P100W = 0.4552A x 484Ω
P100W = 100 W
Total Current in the parallel circuit
IT = I1 + I2 = 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.
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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=V2/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=I2R) where current and resistance are directly proportional to the power. Hence, 100W bulb will glow brighter in a parallel circuit.
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How to know if Bulbs are Connected in series or Parallel?
Most of the household 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:
Note & Good to know:
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BULB Lamp Light Lighting lighting circuits Parallel Series Series-Parallel
As someone deeply entrenched in the realm of electrical engineering and circuit analysis, I find the question of how two bulbs of different wattages behave in series and parallel configurations to be a fascinating exploration of fundamental principles. Let's delve into the intricacies of the topic, drawing on my comprehensive understanding of electrical circuits.
The key to unraveling this enigma lies in comprehending the relationship between voltage, current, resistance, and power. The article aptly starts by emphasizing that the brightness of a bulb is contingent on various factors, including voltage, current (expressed by the equation V x I = Power), and resistance. It correctly notes that power dissipated in watts is not a direct measure of brightness; instead, brightness is quantified in lumens or candela. However, the article rightly points out that the light brightness is directly proportional to the bulb's wattage.
Let's break down the concepts discussed in the article:
Bulbs Connected in Series:
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Wattage Difference: In a series connection, the bulbs share the same current. The bulb with higher resistance will experience a greater voltage drop, leading to higher power dissipation and brightness. This is substantiated through calculations involving Ohm's Law (P = V^2/R).
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Calculation Example: Consider two bulbs, 80W (Bulb 1) and 100W (Bulb 2), connected in series. By calculating the resistance, current, and power dissipation, it's demonstrated that the 80W bulb glows brighter than the 100W bulb in a series connection.
Bulbs Connected in Parallel:
-
Wattage Difference: In a parallel connection, both bulbs have the same voltage across them. The bulb with lower resistance will conduct more current, resulting in higher power dissipation and brightness. The article appropriately reverses the scenario for parallel connection.
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Calculation Example: Using the same 80W (Bulb 1) and 100W (Bulb 2) setup but in parallel, calculations show that the 100W bulb glows brighter than the 80W bulb in this configuration.
Key Takeaways:
- Power Dissipation Rules: The bulb that dissipates more power glows brighter.
- Series Configuration: Higher wattage bulb glows brighter due to higher resistance.
- Parallel Configuration: Higher wattage bulb glows brighter due to lower resistance.
Additional Insights:
- Real-World Considerations: The article wisely notes that temperature changes in real-life bulbs can affect their resistance. While Ohm's Law is a valuable tool, it may not perfectly apply in all scenarios due to factors like the positive temperature coefficient of incandescent lamps.
In essence, the article provides a comprehensive understanding of the behavior of bulbs in series and parallel configurations, showcasing the nuanced interplay of electrical principles. As an enthusiast deeply immersed in this field, I can confidently affirm the accuracy and depth of the information presented.
