Science (086) — AI-generated practice question

Bulbs glow brighter in parallel arrangement.

Equivalent Resistance: In series, the equivalent resistance $R_s = R_1 + R_2 + R_3 = 3R$, which is higher than a single bulb's resistance. In parallel, $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R}$, giving $R_p = R/3$, which is much lower.

Current Distribution: Since the same battery is used, the lower equivalent resistance in parallel allows a greater total current to flow. Each bulb in parallel receives the full battery voltage, drawing more current individually. In series, the same small current flows through all bulbs and the voltage is divided among them.

Therefore, each bulb receives more power ($P = I^2R$ or $P = V^2/R$) in the parallel arrangement and glows brighter.

Source: Electricity, sections 11.6.1 and 11.6.2

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• Examiners award marks for three clear points: (1) comparison of equivalent resistance, (2) current distribution/behavior, and (3) conclusion about brightness.
• Always mention both series formula ($R_s = R_1+R_2+R_3$) and parallel formula to show higher vs. lower resistance — this directly earns marks.
• The key insight: in parallel, each bulb gets the full terminal voltage; in series, voltage is shared, so current and power per bulb are less.
• Quoting $P = V^2/R$ or $P = I^2R$ as justification for brightness is a valued finishing touch.