In a combined circuit containing a bulb in series with a parallel branch, which bulb tends to be brighter according to the source?

In a combined circuit containing a bulb in series with a parallel branch, which bulb tends to be brighter according to the source?

• A. The bulb in series is brightest; parallel bulbs have equal brightness and are dimmer than the series bulb.
• B. The parallel bulbs are brightest.
• C. All bulbs have the same brightness.
• D. The series bulb is dimmer than the parallel bulbs.

Answer: (A)

Brightness corresponds to how much power a bulb dissipates. In a circuit where one bulb is in series with a parallel branch, the current through the series bulb is the same as the total current from the source, while the bulbs in the parallel branch each see the same voltage across them (the branch voltage) but share the current. If we assume the series bulb has resistance R and the two parallel bulbs are identical with resistance R each, the parallel combination has equivalent resistance R/2. The total resistance is R + R/2 = 3R/2, so the source current is I = V / (3R/2) = 2V/(3R). The power in the series bulb is P_series = I^2 R = 4V^2/(9R). The voltage drop across the series bulb is V_series = IR = 2V/3, leaving V_branch = V − V_series = V/3 across the parallel bulbs. Each parallel bulb then has power P_each = (V_branch)^2 / R = V^2/(9R). Since P_series = 4V^2/(9R) while each parallel bulb has V^2/(9R), the series bulb is brighter, and the two parallel bulbs have equal brightness to each other but are dimmer than the series bulb. So the best answer is that the bulb in series is brightest; the parallel bulbs have equal brightness and are dimmer than the series bulb.

Brightness corresponds to how much power a bulb dissipates. In a circuit where one bulb is in series with a parallel branch, the current through the series bulb is the same as the total current from the source, while the bulbs in the parallel branch each see the same voltage across them (the branch voltage) but share the current.

If we assume the series bulb has resistance R and the two parallel bulbs are identical with resistance R each, the parallel combination has equivalent resistance R/2. The total resistance is R + R/2 = 3R/2, so the source current is I = V / (3R/2) = 2V/(3R). The power in the series bulb is P_series = I^2 R = 4V^2/(9R). The voltage drop across the series bulb is V_series = IR = 2V/3, leaving V_branch = V − V_series = V/3 across the parallel bulbs. Each parallel bulb then has power P_each = (V_branch)^2 / R = V^2/(9R). Since P_series = 4V^2/(9R) while each parallel bulb has V^2/(9R), the series bulb is brighter, and the two parallel bulbs have equal brightness to each other but are dimmer than the series bulb.

So the best answer is that the bulb in series is brightest; the parallel bulbs have equal brightness and are dimmer than the series bulb.