# Why Does the High-Wattage Bulb Glow Dimmer in a Series Circuit?

## 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?

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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: I
_{T}= I_{1}= I_{2}= I_{3}…I_{n} - Voltage in series circuit: V
_{T}= V_{1}+ V_{2}+ V_{3}…+ V_{n} - Resistance in series circuit: R
_{T}= R_{1}+ R_{2}+ R_{3}…+ R_{n}

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**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)*

**➺**R_{10W}= V^{2}/ P_{10}**➺**R_{10W }= 230^{2}/ 10W**➺**R_{10W }= 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)**

**➺**R_{50W}= V^{2}/ P_{50}**➺**R_{50W }= 230^{2}/ 50W**➺**R_{50W }= 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 ÷ R_{Total}**➺**I = V ÷ (R_{10W}+ R_{50W}+ R_{10W})**➺**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 = I^{2}R).

*Power Dissipated by 10W Bulb (Each)*

**➺**P_{10W}= I^{2}R_{10W}**➺**P_{10W }= (19.76mA)^{2}x 5,290Ω**➺**P_{10W }= 2 Watts.

*Power Dissipated by 50W Bulb*

Similarly, the power dissipated by the 50W bulb connected in the middle of the series circuit

**➺**P_{50W}= I^{2}R_{50W}**➺**P_{50W }= (19.76mA)^{2}x 1,058Ω**➺**P_{50W }= 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.

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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 = R
_{T}= 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 = V
_{Drop}= V_{1}+ V_{2}+ V_{3}= 230V - Total Current = I
_{T}= 19.76 mA

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**Good to know:**The scenario is totally different and opposite when all the above bulbs, having the same rating, are connected in a parallel circuit. In a parallel arrangement, the high-wattage bulb will glow brighter, and the low-wattage bulbs will glow dimly.

Based on the above calculations, we now understand why the low-wattage rated bulbs (10W) glow brighter and the high-wattage rated bulb glows dimmer when connected in a series circuit. This is because the same current flows through all bulbs, while the wattage and resistance values are different. The low-wattage bulbs have high resistance and dissipate more power hence, they glow brighter. While the high-wattage bulb has low resistance and dissipates less power, hence, it glows dimly.

**Good to Know:**That is the objection to connecting lighting points in series. Instead, a parallel connection is preferred for connecting lighting fixtures because it has some advantages over a series circuit.

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