Question

7 while working on her science fair project, venus connected a battery to a circuit that contained a light bulb. venus decided to change the battery to a higher voltage, but she kept the same light bulb. what will happen to the brightness of the bulb?
a the brightness will decrease because the current will decrease
b the brightness will decrease because the current will increase
c the brightness will increase because the current will decrease
d the brightness will increase because the current will increase

Explanation:

Step1: Recall Ohm's Law

Ohm's Law is \(V = IR\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance. The resistance \(R\) of the light - bulb remains the same (since it's the same bulb). If \(V\) (voltage) increases (as the battery voltage is increased) and \(R\) is constant, then from \(I=\frac{V}{R}\), the current \(I\) will increase.

Step2: Relate current to brightness

The brightness of a light - bulb is related to the power \(P\) dissipated in the bulb. The power formula for a resistor (light - bulb) is \(P = I^{2}R\) (or \(P=\frac{V^{2}}{R}\)). Since \(R\) is constant and \(I\) (or \(V\)) increases, the power \(P\) increases. A higher power means a brighter bulb.

Answer:

D. The brightness will increase because the current will increase