Understanding Bulb Brightness in Different Circuit Arrangements

This explanation details how the arrangement of identical bulbs in an electrical circuit affects their brightness. We will analyze four different configurations: all bulbs in series, all bulbs in parallel, and two mixed configurations. The brightness of a bulb depends on the power it dissipates, calculated as $P = V^2 / R$ or $P = I^2 \times R$. For identical bulbs, the resistance ($R$) is the same.

Key Concepts for Brightness

• Brightness and Power: The brightness of a bulb is directly related to the electrical power it dissipates. More power means a brighter glow.
• Power Formulas: Power ($P$) can be calculated using the voltage ($V$) across the bulb and its resistance ($R$) as $P = V^2 / R$, or using the current ($I$) through the bulb and its resistance ($R$) as $P = I^2 \times R$.
• Identical Bulbs: Since the bulbs are identical, they all have the same resistance, let's call it $R$.
• Power Source: Assume the circuit is powered by a source with a constant voltage, $V$.

Analysis of Circuit Arrangements

Arrangement A: All Bulbs in Series

In this arrangement, the three identical bulbs are connected end-to-end, forming a single path for the current.

• Total Resistance: The total resistance of the circuit is the sum of individual resistances: $R_{total\_A} = R + R + R = 3R$.
• Total Current: The current flowing through the circuit is determined by Ohm's Law ($I = V/R_{total}$): $I_A = V / (3R)$.
• Voltage Across Each Bulb: Since the current is the same through all bulbs, the voltage drops across each bulb are equal: $V_{bulb\_A} = I_A \times R = (V / (3R)) \times R = V/3$.
• Power Dissipated by Each Bulb: Using $P = V^2 / R$: $P_A = (V/3)^2 / R = V^2 / (9R)$.

In this configuration, each bulb receives only one-third of the total voltage, resulting in low power dissipation and dim light.

Arrangement B: All Bulbs in Parallel

Here, each bulb is connected directly across the power source.

• Total Resistance: The reciprocal of the total resistance is the sum of the reciprocals of individual resistances: $1/R_{total\_B} = 1/R + 1/R + 1/R = 3/R$. Thus, $R_{total\_B} = R/3$.
• Voltage Across Each Bulb: In a parallel arrangement, each component receives the full voltage of the source: $V_{bulb\_B} = V$.
• Power Dissipated by Each Bulb: Using $P = V^2 / R$: $P_B = V^2 / R$.

With each bulb receiving the full source voltage, the power dissipation is maximized, making the bulbs glow the brightest in this arrangement.

Arrangement C: Two Bulbs in Parallel Connected in Series with the Third Bulb

This is a mixed circuit. A pair of bulbs are connected in parallel, and this combination is then connected in series with the third bulb.

• Resistance of the Parallel Pair: $R_{parallel} = (R \times R) / (R + R) = R/2$.
• Total Resistance: The total resistance is the sum of the parallel resistance and the third bulb's resistance: $R_{total\_C} = R/2 + R = 3R/2$.
• Total Current: $I_C = V / R_{total\_C} = V / (3R/2) = 2V / (3R)$. This is the current through the third (series) bulb.
• Power Dissipated by the Series Bulb: $P_{series\_bulb} = I_C^2 \times R = (2V / (3R))^2 \times R = (4V^2 / (9R^2)) \times R = 4V^2 / (9R)$.
• Voltage Across the Parallel Pair: $V_{parallel\_pair} = I_C \times R_{parallel} = (2V / (3R)) \times (R/2) = V/3$.
• Voltage Across Each Parallel Bulb: Since they are in parallel, the voltage across each is the same: $V_{parallel\_bulbs} = V_{parallel\_pair} = V/3$.
• Power Dissipated by Each Parallel Bulb: $P_{parallel\_bulbs} = (V/3)^2 / R = V^2 / (9R)$.

In this case, the series bulb glows brighter than the bulbs in Arrangement A, but dimmer than the bulbs in Arrangement B. The parallel bulbs glow dimly.

Arrangement D: Two Bulbs in Series Connected in Parallel with the Third Bulb

This is another mixed circuit. A pair of bulbs are connected in series, and this combination is then connected in parallel with the third bulb.

• Resistance of the Series Pair: $R_{series} = R + R = 2R$.
• Circuit Analysis: The branch with the series pair (resistance $2R$) is in parallel with the branch containing the single bulb (resistance $R$). Both branches are connected across the voltage source $V$.
• Current and Power for the Single Bulb Branch: The voltage across this branch is $V$. Current $I_1 = V/R$. Power $P_1 = V^2 / R$.
• Current and Power for the Series Pair Branch: The voltage across this branch is also $V$. The total resistance of this branch is $2R$. Current $I_2 = V / (2R)$.
• Power Dissipated by Each Bulb in the Series Pair: Since the current $I_2$ flows through both bulbs in the series pair, the power dissipated by each is $P_{2a} = P_{2b} = I_2^2 \times R = (V / (2R))^2 \times R = (V^2 / (4R^2)) \times R = V^2 / (4R)$.

In this arrangement, the single bulb glows brightly (same power as in Arrangement B). However, the two bulbs in the series pair glow less brightly than the single bulb, dissipating $V^2 / (4R)$ each.

Comparing Brightness Across Arrangements

Let's compare the power dissipated by individual bulbs in each arrangement to determine the brightest glow:

• Arrangement A (All Series): $P_{bulb} = V^2 / (9R)$
• Arrangement B (All Parallel): $P_{bulb} = V^2 / R$
• Arrangement C (Mixed): Max power is for the series bulb, $P_{series\_bulb} = 4V^2 / (9R)$. The parallel bulbs have $P_{parallel\_bulbs} = V^2 / (9R)$.
• Arrangement D (Mixed): Max power is for the single parallel bulb, $P_1 = V^2 / R$. The series bulbs have $P_{2a} = P_{2b} = V^2 / (4R)$.

The highest power dissipation for an individual bulb occurs in Arrangement B (all parallel) and for the single bulb in Arrangement D. Both achieve $P = V^2 / R$. However, Arrangement B has all three bulbs glowing at this maximum power. Arrangement D only has one bulb glowing at $V^2 / R$, while the other two glow less brightly ($V^2 / (4R)$).

Therefore, the arrangement where the bulbs glow the brightest is Arrangement B, as all bulbs simultaneously achieve the maximum possible power dissipation ($V^2 / R$) for the given conditions.

Conclusion

The arrangement where all bulbs are connected in parallel (Arrangement B) results in the brightest glow because each bulb receives the full voltage from the power source. This maximizes the power dissipated by each identical bulb compared to series or mixed configurations where voltage is divided.