Which statement about brightness in a mixed circuit is true according to the material?

Which statement about brightness in a mixed circuit is true according to the material?

• A. The series bulb is brighter than the parallel bulbs
• B. The series bulb is dimmer than the parallel bulbs
• C. All bulbs have equal brightness
• D. Brightness cannot be predicted

Answer: (A)

Brightness comes from how much electrical power a bulb dissipates. With identical bulbs in a mixed circuit where one bulb is in series with two identical bulbs in parallel, the parallel pair looks like a single resistor of resistance R/2 to the rest of the circuit. The series bulb (resistance R) and this parallel block are in series, so the total resistance is R + R/2 = 3R/2. The current from the source is I = V / (3R/2) = 2V/(3R). The series bulb carries this same current, so its power is P_series = I^2 R = (4/9) V^2 / R. The voltage across the parallel block is V_parallel = I × (R/2) = (2V/(3R)) × (R/2) = V/3. Each parallel bulb has V/3 across it, giving power P_parallel = (V/3)^2 / R = V^2/(9R) for each bulb. Compare the powers: the series bulb dissipates (4/9) V^2/R, while each parallel bulb dissipates (1/9) V^2/R. The series bulb is brighter because it dissipates four times as much power as each parallel bulb. The two parallel bulbs together would yield 2 × (V^2/(9R)) = 2V^2/(9R), which is still less than the series bulb’s power, so the single series bulb appears brighter.

Brightness comes from how much electrical power a bulb dissipates. With identical bulbs in a mixed circuit where one bulb is in series with two identical bulbs in parallel, the parallel pair looks like a single resistor of resistance R/2 to the rest of the circuit. The series bulb (resistance R) and this parallel block are in series, so the total resistance is R + R/2 = 3R/2. The current from the source is I = V / (3R/2) = 2V/(3R). The series bulb carries this same current, so its power is P_series = I^2 R = (4/9) V^2 / R.

The voltage across the parallel block is V_parallel = I × (R/2) = (2V/(3R)) × (R/2) = V/3. Each parallel bulb has V/3 across it, giving power P_parallel = (V/3)^2 / R = V^2/(9R) for each bulb.

Compare the powers: the series bulb dissipates (4/9) V^2/R, while each parallel bulb dissipates (1/9) V^2/R. The series bulb is brighter because it dissipates four times as much power as each parallel bulb. The two parallel bulbs together would yield 2 × (V^2/(9R)) = 2V^2/(9R), which is still less than the series bulb’s power, so the single series bulb appears brighter.