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

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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?

The primary reason for this scenario is the fundamental principle that a bulb dissipating more power will glow brighter, regardless of whether it is connected in a series or parallel circuit. In other words, a bulb with higher resistance will glow brighter, while a bulb with lower resistance will glow dimmer. This is because the bulb with lower resistance dissipates less power. The brightness of a bulb is directly proportional to the power it dissipates, as more power results in more light output.

Now, we will calculate the current flowing through each bulb and determine the resistance of each one because the brightness of a light bulb depends on the current, voltage, and its resistance. These values are important because the brightness of the bulb is determined by the power it dissipates, not simply by its rated power.

Since this is a parallel circuit, the voltage across each branch remains the same, while the total current is the sum of the currents through each branch. Additionally, the total resistance of multiple resistors in parallel is always less than the smallest individual resistance in the circuit.

Current in parallel circuit: I_T + I₁ + I₂ + I₃ … + I_n
Voltage in parallel circuit: V_T = V₁ = V₂ = V₃ … = V_n
Resistance in parallel circuit: R_T = 1/R₁ + 1/R₂ + 1/R₃ …+ 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 = V² / P_5W
➺ R_5W= 230² / 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)

➺ R₂₅_W = V² / P₂₅
➺ R₂₅_W= 230² / 25W
➺ R₂₅_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).

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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 = V² ÷ R).

Power Dissipated by 5W Bulbs (Each)

➺ P_5W = V² ÷ 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 = V² ÷ R_25W
➺ P_25W= (230V)² x 2,116Ω
➺ P_25W= 25 Watts.

Now, we observe that the 25-watt bulb, which is connected in the middle of the parallel circuit, is consuming more power (25 watts) compared to the first and last bulbs, each of which consumes 5 watts. This is why the middle bulb, having a higher wattage rating, glows brighter than the other two lower-wattage bulbs, which glow dimmer.

Alternatively, you can calculate the total resistance of the bulbs using the formula for equivalent resistance in a parallel circuit. In this case, the equivalent resistance is 1,511.43 ohms. After finding the total resistance, you can calculate the voltage drop across each bulb by multiplying the current flowing through each bulb by its corresponding resistance. Use a calculator to perform these calculations accurately.

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 = I₁ + I₂ + I₃ = 152.2 mA
Total Voltage Drops = V_Drop = 230V

Good to Know: The scenario is completely different and opposite when all the above bulbs, having the same ratings, are connected in a series circuit. In a series arrangement, the high-wattage bulb will glow dimmer, while the low-wattage bulbs will glow brighter..

Based on the above calculations, we now understand why the low-wattage bulbs (5W) glow dimmer, while the high-wattage bulb glows brighter in a parallel circuit. This happens because the same voltage is applied across each bulb, but the wattage ratings and resistance values differ.

The high-wattage bulb has a lower resistance, allowing more current to flow through it, which results in higher power dissipation and brighter illumination. In contrast, the low-wattage bulbs have higher resistance, so less current flows through them, resulting in lower power dissipation and dimmer light.

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.