# Why Does the High-Wattage Bulb Glow Brighter in a Parallel Circuit?

## Why Does a High-Voltage Bulb Glow Brightly When Connected in a Parallel Circuit, While a Low-Voltage Bulb Glows Dimly in the Same Circuit?

Three light bulbs are connected in a parallel circuit across the source voltage of 230V AC supply.

- The rating of the first bulb is 5 Watts
- The rating of the middle bulb is 25 Watts.
- The rating of the last bulb is 5 watts.

Now, why does the middle bulb (high wattage 25 watts) connected in the parallel circuit glow brighter than the first and last bulbs (each rated at 5 watts) connected in the same circuit and glowing dimmer?

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To understand the reason behind this scenario, you need to know 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. We are interested in these values 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 parable circuit, we know that in a parallel circuit, the voltage is the same across each point while the current is additive. The total resistance of multiple resistors in a parallel circuit is less than the individual resistance.

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

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**Finding Total Resistances of the Bulbs**

As there will be the same voltage level, i.e., 230V, across each bulb in the parallel circuit, 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 5W Bulb (Each)*

**➺**R_{5W}= V^{2}/ P_{5W}**➺**R_{5W}_{ }= 230^{2}/ 5W**➺**R_{5W}_{ }= 10,580 Ω

This is the resistance value for both the first and last bulbs, which are rated at 5W. Let’s find the resistance of the bulb connected in the middle, which has a high wattage rating, e.g., 25W.

**Resistance of 25W Bulb (Connected in the Middle)**

**➺**R_{25}_{W}= V^{2}/ P_{25}**➺**R_{25}_{W }= 230^{2}/ 25W**➺**R_{25}_{W }= 2,116 Ω

We see that the resistance of the high-wattage (25W) rated bulb connected in the middle of the parallel circuit is lower than the resistance of the low-wattage bulbs (5W).

**Finding Power Dissipation of the Bulbs**

As we know that the same voltage (e.g. 230V) is available across each light bulb, 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 = V^{2} ÷ R).

*Power Dissipated by 5W Bulbs (Each)*

**➺**P_{5W}= V^{2}÷ R_{5W}**➺**P_{5W }= (230V)^{2}x 10,580Ω**➺**P_{5W }= 5 Watts.

*Power Dissipated by 25W Bulb*

Similarly, the power dissipated by the 25W bulb connected in the middle of the parallel circuit

**➺**P_{25W}= V^{2}÷ R_{25W}**➺**P_{25W}_{ }= (230V)^{2}x 2,116Ω**➺**P_{25W}_{ }= 25 Watts.

Now, we see that the 25W bulb connected in the middle of the parallel circuit is consuming more power (25 Watts) compared to the first and last bulbs (each rated at 5W), which consume 5 Watts each. That’s why the middle bulb (having high wattage rating) connected in the parallel circuit, is glowing brighter compared to the other low-wattage bulbs, which glow dimmer.

Alternatively, you can calculate the total resistance of bulbs (using the equivalent resistance formula of parallel circuit), so the equivalent resistance will be 1,511.43 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}= 1,511.43 Ohms - Current flowing in 5-W bulb (each) = 21.74 mA
- Current flowing in 50-W bulb = 108.8 mA
- Total Current = I
_{T}= I_{1}+ I_{2}+ I_{3}= 152.2 mA - Total Voltage Drops = V
_{Drop}= 230V

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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 series circuit. In a series arrangement, the high-wattage bulb will glow dimmer, and the low-wattage bulbs will glow brighter.

Based on the above calculations, we now understand why the low-wattage rated bulbs (5W) glow dimmer and the high-wattage rated bulb glows brighter when connected in a parallel circuit. This is because the same voltage are available across each bulb, while the wattage and resistance values are different. The high-wattage bulb has low resistance, hence more current flows in it and dissipate more power, hence, it glows brightly. On the other hand, the low-wattage bulbs have high resistances and less current flows through them, hence both dissipate less power and glows dimly.

**Good to Know:**The objection to connecting lighting points in series is that a parallel connection is preferred for connecting lighting fixtures, as it offers several advantages over a series circuit.

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