Question

fill in the blank:
\if we shorten the
resistance wire in a circuit,
the bulbs will get ____\
choose your answer by clicking one of the
options, then click \ok\
light all 5 stars for the bonus!
section 4 // question 5
changing circuits
help

Explanation:

Step1: Recall Ohm's Law

Ohm's Law is \(V = IR\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance. In a circuit, if the voltage source (assuming it's constant, like a battery) remains the same, and we change the resistance. The power dissipated in a bulb (which determines its brightness) is \(P=I^{2}R\) (also \(P = \frac{V^{2}}{R}\) when \(V\) is constant).

Step2: Analyze the effect of shortening the wire

When we shorten the wire in a circuit, the resistance \(R\) of the wire (assuming the wire has some resistance, and it's in series with the bulb) decreases. Using \(I=\frac{V}{R_{total}}\) (where \(R_{total}=R_{wire}+R_{bulb}\)), if \(R_{wire}\) decreases, \(R_{total}\) decreases. Since \(V\) is constant (from the power source), \(I\) (the current in the circuit) increases. Using \(P = I^{2}R_{bulb}\) (assuming \(R_{bulb}\) is constant for a given bulb), as \(I\) increases, \(P\) (the power consumed by the bulb) increases. A higher - power consumption means the bulb is brighter.

Answer:

a. Brighter