Question

1. you instead want to make sure the battery for your string of lights will last as long as possible. a battery w last longer if it powers a circuit with low current. how could you hook up a battery and 2 light bulbs so the least amount of current flows through the battery? use the measurement tools in the simulation to check your design.

Explanation:

To minimize the current through the battery (and thus make it last longer), we use the relationship between resistance, voltage, and current ($I = \frac{V}{R}$ from Ohm's Law). When components are in series, the total resistance $R_{total}$ is the sum of individual resistances ($R_{total}=R_1 + R_2$ for two resistors). When in parallel, $R_{total}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}$, which is less than either $R_1$ or $R_2$. Since $I=\frac{V}{R}$, a larger total resistance will result in a smaller current (as voltage $V$ from the battery is constant). So, hooking the two light bulbs in series will increase the total resistance, leading to the least current through the battery. We can verify this using an ammeter (measurement tool) to measure the current in series and parallel circuits; the series circuit will show a lower current reading.

Answer:

Hook the two light bulbs in series with the battery. This increases the total resistance of the circuit (since $R_{\text{total}} = R_1 + R_2$ for series resistors), and by Ohm's Law ($I=\frac{V}{R}$), a larger total resistance with a constant battery voltage $V$ results in a smaller current $I$ flowing through the battery, which helps the battery last longer.