Question

modern nimh rechargeable batteries of size aa have a maximum charge capacity of 2300 mah and an emf equal to 1.20 v. a flashlight has three of these batteries connected in series. the 3.6 v bulb in the flashlight draws a current of 0.3 a. assume that both the emf and the current do not change with time. part a calculate how long the flashlight will operate using these rechargeable batteries.

Explanation:

Step1: Calculate total charge

The capacity of one battery is 2300 mAh. Since there are 3 batteries in series, the total charge $Q$ (in mAh) remains the same as that of a single - battery's capacity in this case (because in series connection, charge capacity doesn't add up in the way voltage does), so $Q = 2300\ \text{mAh}\times3=6900\ \text{mAh}$. Convert it to A - s. $1\ \text{mAh}=1\times10^{- 3}\ \text{A}\times3600\ \text{s} = 3.6\ \text{A}\cdot\text{s}$, so $Q = 6900\times10^{-3}\ \text{A}\times3600\ \text{s}=24840\ \text{A}\cdot\text{s}$.

Step2: Use the formula $Q = It$

We know that the current $I = 0.3\ \text{A}$, and from the formula $Q = It$, we can solve for time $t$. Rearranging the formula gives $t=\frac{Q}{I}$. Substitute $Q = 24840\ \text{A}\cdot\text{s}$ and $I = 0.3\ \text{A}$ into the formula, we get $t=\frac{24840}{0.3}\ \text{s}=82800\ \text{s}$.

Step3: Convert time to hours

Since $1\ \text{hour}=3600\ \text{s}$, then $t=\frac{82800}{3600}\ \text{h}=23\ \text{h}$.

Answer:

23