Question

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Explanation:

Step1: Relate brightness to power

The brightness of a bulb depends on the power it dissipates, given by $P = \frac{V^2}{R}$ (where $V$ is voltage across the bulb, $R$ is bulb resistance, assumed constant) or $P=I^2R$ (where $I$ is current in the circuit). Lower power means dimmer light.

Step2: Analyze option a (lower battery voltage)

If battery voltage $V_{batt}$ decreases, the voltage across the bulb $V$ decreases. Using $P = \frac{V^2}{R}$, lower $V$ reduces $P$, so bulb dims.

Step3: Analyze option b (higher battery voltage)

Increasing $V_{batt}$ increases $V$ across the bulb. $P = \frac{V^2}{R}$ increases, so bulb gets brighter.

Step4: Analyze option c (lower circuit resistance)

Total circuit resistance $R_{total}$ decreases. By Ohm's law $I = \frac{V_{batt}}{R_{total}}$, current $I$ increases. Using $P=I^2R$, higher $I$ increases $P$, bulb gets brighter.

Step5: Analyze option d (higher circuit resistance)

$R_{total}$ increases, so $I = \frac{V_{batt}}{R_{total}}$ decreases. $P=I^2R$ decreases, so bulb dims.

Step6: Analyze option e (open the circuit)

An open circuit means no current flows ($I=0$). $P=I^2R=0$, so the bulb stops shining (which is less bright than a lit bulb).

Answer:

a. decrease the voltage of the battery
d. increase the resistance of the circuit
e. open the circuit