Why Does a High-Wattage Bulb Glow Dimly When Connected in a Series Circuit, While a Low-Voltage Bulb Glows Brightly in the Same Circuit?
Three light bulbs are connected in a series circuit which is powered by the source voltage of 230V AC supply.
- The rating of the first bulb is 10 Watts
- The rating of the middle bulb is 50 Watts.
- The rating of the last bulb is 10 watts.
Now, why is the high-wattage (50 watts) bulb connected in the middle of a series circuit glowing dimly compared to the first and last bulbs, each rated at 10 watts, which are glowing brightly?
To understand the reason behind this scenario, you need to grasp the basic principle that a bulb dissipating more power will glow brighter, whether it is connected in a series or parallel circuit. In other words, a bulb with high resistance will glow brighter, while a bulb with low resistance will glow dimmer. This is because the bulb with lower resistance dissipates less power, where the brightness of the bulb is directly proportional to the light brightness.
Now, we will find the current flowing through and the resistance of each bulb because the brightness of the light bulb depends on the current, voltage, and its resistance. This is because the brightness of the light depends on the power dissipated by the light bulb, not the rating of the light bulb.
As it is a series circuit, we know that in a series circuit, the current is the same at each point while the voltage and resistances are additive, i.e.
- Current in series circuit: IT = I1 = I2 = I3 …In
- Voltage in series circuit: VT = V1 + V2 + V3 …+ Vn
- Resistance in series circuit: RT = R1 + R2 + R3 …+ Rn
Determining the Total Resistance of the Lights
As the same current flows through all three bulbs connected in series, we can use Ohm’s law to determine the resistance of each bulb, which will allow us to calculate the power dissipation by the light bulb (P = V×I). Let’s find the resistance of each bulb using Ohm’s Law.
Resistance of 10W Bulb (Each)
- ➺ R10W = V2 / P10
- ➺ R10W = 2302 / 10W
- ➺ R10W = 5,290 Ω
This is the resistance value for both the first and last bulbs, which are rated at 10W. Let’s find the resistance of the bulb connected in the middle, which has a high wattage rating, e.g., 50W.
Resistance of 50W bulb (Connected in the Middle)
- ➺ R50W = V2 / P50
- ➺ R50W = 2302 / 50W
- ➺ R50W = 1,058 Ω
We see that the resistance of the high-wattage (50W) rated bulb connected in the middle of the series circuit is lower than the resistance of the low-wattage bulbs (10W).
Calculating the Current Flowing in the Circuit
Now, let’s determine the total current flowing in the series circuit using Ohm’s law.
- ➺ I = V ÷ RTotal
- ➺ I = V ÷ (R10W + R50W + R10W)
- ➺ I = 230V ÷ (5,290Ω + 1,058Ω + 5,290Ω)
- ➺ I = 19.76 mA.
Calculating the Power Dissipation of the Bulbs
Hence, the power dissipated by the lower wattage bulb (either the first one or the last one, each having a 5-watt rating) can be calculated using the power dissipation formula (P = I2R).
Power Dissipated by 10W Bulb (Each)
- ➺ P10W = I2R10W
- ➺ P10W = (19.76mA)2 x 5,290Ω
- ➺ P10W = 2 Watts.
Power Dissipated by 50W Bulb
Similarly, the power dissipated by the 50W bulb connected in the middle of the series circuit
- ➺ P50W = I2R50W
- ➺ P50W = (19.76mA)2 x 1,058Ω
- ➺ P50W = 0.41 Watts.
Now, we see that the 50W bulb connected in the middle of the series circuit is consuming less power (0.41 Watts) compared to the first and last bulbs (each rated at 10W), which consume 2 Watts each. That’s why the middle bulb connected in the series circuit, even though it has a high-wattage rating, is glowing very dimly compared to the other low-wattage bulbs, which glow brighter.
Alternatively, you can calculate the total resistance (by adding the resistance of each bulb), so the equivalent resistance will be 11,638 ohms. Then, you can calculate the voltage drop across each bulb by multiplying the current flowing through the bulb by the related resistance of that bulb. Use a calculator for these calculations.
- Total Resistance = RT = 11,638 Ohms
- Voltage drop across 5-W bulb (each) = 104.55 V
- Voltage drop across 50-W bulb = 20.9 V
- Total Voltage drop = VDrop = V1 + V2 + V3 = 230V
- Total Current = IT = 19.76 mA
FAQ
Why do bulbs get dimmer in a series circuit?
Adding another lamp to a series circuit increases the resistance, so the current is lower and the lamps are dimmer. Current is not used up by the components in a circuit
Why do bulbs glow brighter in series?
The bulb glows brighter and brighter as we keep on adding more cells in series. This happens because more chemical energy is converted to electrical energy when two cells are used. Consequently, the flow of electric current is greater
What makes a bulb glow brighter?
When more power is consumed by the bulb, the heat energy generated will be greater, thus allowing the filament to glow brighter
Why do bulbs not get dimmer in parallel?
Two bulbs in a simple parallel circuit each enjoy the full voltage of the battery. This is why the bulbs in the parallel circuit will be brighter than those in the series circuit. Another advantage to the parallel circuit is that if one loop is disconnected, then the other remains powered
Why do LEDs in series get dimmer?
When connecting multiple strips, the voltage drop can cause LEDs at the end of the strip to appear dimmer than those at the beginning. The longer the LED strip, the more significant the voltage drop will be. Voltage drop describes the decrease in voltage level along the length of a circuit due to internal resistance
What happens if a 50w and 100w bulb are connected in series?
The 50w bulb will glow more, because the same current flow through the bulbs in series. Power is I2R, so the larger R will dissipate more power
Which bulb will glow brighter, 60w or 100w in series?
The resistance of the bulb is given by R=V2/P. So the resistance of 60 W bulb is more than the resistance of 100 W bulb. When they are connected in series the current through both bulbs is same. Hence 60 W bulb will be more brighter because P = I2R
Why does the brightness change in a series circuit but not in a parallel circuit?
When in series, bulbs become dimmer as the potential difference is shared equally across the bulbs. The current reads the same for each component. In parallel, each branch shows the same potential difference, so the bulbs on one branch will have the same relative brightness
