Question

part two: electromagnetic induction

1. a compact fluorescent light bulb draws a current of 0.10 a for one hour.

a how much charge flows through the bulb in one hour? (2 marks)

Explanation:

Step1: Recall the formula for charge

The formula relating current (\(I\)), charge (\(Q\)) and time (\(t\)) is \(I=\frac{Q}{t}\), which can be rearranged to \(Q = I\times t\).

Step2: Convert time to seconds

We know that \(1\) hour \(= 3600\) seconds. The current \(I = 0.10\space A\) and time \(t=3600\space s\).

Step3: Calculate the charge

Substitute the values of \(I\) and \(t\) into the formula \(Q = I\times t\). So \(Q=0.10\space A\times3600\space s\).
\(Q = 360\space C\) (since \(1\space A = 1\space C/s\), so \(A\times s = C\)).

Answer:

The charge that flows through the bulb in one hour is \(\boldsymbol{360\space C}\) (coulombs).