Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are connected in series across a 120 V line. Afterwards, the two light bulbs are connected in parallel across the 120 V line. In each situation, which of the two bulbs glows the brightest?

Verified step by step guidance

1. Understand the concept of brightness in light bulbs: Brightness is related to the power consumed by the bulb, which can be calculated using the formula \( P = \frac{V^2}{R} \) for series and parallel circuits.
2. Calculate the equivalent resistance for the bulbs in series: When resistors are in series, the total resistance \( R_{total} \) is the sum of individual resistances. So, \( R_{total} = R_1 + R_2 = 400\,\Omega + 800\,\Omega = 1200\,\Omega \).
3. Determine the current through the bulbs in series: Use Ohm's Law \( I = \frac{V}{R_{total}} \) to find the current. With \( V = 120\,V \) and \( R_{total} = 1200\,\Omega \), calculate \( I \). Then, use \( P = I^2 \times R \) to find the power for each bulb.
4. Calculate the equivalent resistance for the bulbs in parallel: For parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances. \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \). Solve for \( R_{total} \).
5. Determine the power for each bulb in parallel: Since the voltage across each bulb is the same (120 V), use \( P = \frac{V^2}{R} \) for each bulb to find the power. Compare the power values to determine which bulb glows the brightest in parallel.

Key Concepts

Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points, represented by the formula V = IR, where V is voltage, I is current, and R is resistance. This principle helps determine the current flowing through each bulb in both series and parallel configurations.

Series and Parallel Circuits

In a series circuit, components are connected end-to-end, so the same current flows through each component, but the voltage is divided among them. In a parallel circuit, components are connected across the same voltage source, so each component receives the full voltage, but the current is divided based on resistance. Understanding these configurations is crucial for analyzing the brightness of the bulbs.

Power Dissipation in Resistors

Power dissipation in a resistor is given by the formula P = IV, where P is power, I is current, and V is voltage. Alternatively, it can be expressed as P = I²R or P = V²/R. The brightness of a bulb is related to the power it dissipates, so calculating power for each bulb in both series and parallel setups helps determine which bulb glows brightest.