B.5 Current and circuits

Given: $I=0.40\text{ A}$, $\Delta t=5\text{ min}=5\times 60=300\text{ s}$. The formula relating charge, current, and time is $\Delta Q=I\cdot \Delta t$.

Time must be in seconds: $\Delta t=5\times 60=300\text{ s}$. A very common error is leaving time in minutes, which would give an answer 60 times too small.

$\Delta Q=I\cdot \Delta t=0.40\text{ A}\times 300\text{ s}=120\text{ C}$

The total charge transferred is 120 C.

We need the total charge $\Delta Q=I\cdot \Delta t$, with $I=0.40\text{ A}$ and $t=5$ minutes.

2.0 C would result from $0.40\times 5=2.0$, which uses time in minutes instead of seconds. This is a unit error and can be eliminated.

8.0 C does not correspond to any straightforward calculation with the given values. It can be eliminated.

0.080 C would result from dividing current by time ($0.40/5=0.08$), which is incorrect use of the formula. Eliminate this option.

Only 120 C comes from the correct calculation: $0.40\text{ A}\times 300\text{ s}=120\text{ C}$.

Given: $V=4.5\text{ V}$, $I=0.30\text{ A}$. Ohm's Law gives $R=V/I$.

$R=\frac{V}{I}=\frac{4.5\text{ V}}{0.30\text{ A}}=15\text{ Ω}$

The resistance of the bulb filament is 15 Ω.

We need resistance from $R=V/I$ with $V=4.5\text{ V}$ and $I=0.30\text{ A}$.

0.067 Ω comes from $I/V=0.30/4.5\approx 0.067$, which is the reciprocal of resistance (i.e., conductance). This is the wrong formula.

1.35 Ω comes from multiplying $V\times I=4.5\times 0.30=1.35$, which gives power in watts, not resistance. Eliminate.

4.2 Ω is obtained by subtracting the current from the voltage ($4.5-0.30=4.2$), which has no physical basis. Eliminate.

Only 15 Ω is obtained from the correct application of $R=V/I=4.5/0.30=15\text{ Ω}$.