problem 32?\n34. a battery can provide an average of 1.5 a of current for 24 h. what is its ampere - hour…

Explanation:

Step1: Recall the ampere - hour formula

The ampere - hour (Ah) rating of a battery is given by the product of current (I) and time (t), i.e., $Ah = I\times t$.

Step2: Solve problem 34

Given $I = 1.5\ A$ and $t=24\ h$. Using the formula $Ah = I\times t$, we have $Ah=1.5\times24 = 36\ Ah$.

Step3: Solve problem 35

Given $Ah = 80\ Ah$ and $t = 10\ h$. From $Ah = I\times t$, we can find $I=\frac{Ah}{t}$. Substituting the values, $I=\frac{80}{10}=8\ A$.

Step4: Solve problem 36

Given $Ah = 650\ mAh=0.65\ Ah$ and $t = 48\ h$. Using $I=\frac{Ah}{t}$, we get $I=\frac{0.65}{48}\approx0.0135\ A = 13.5\ mA$.

Step5: Solve problem 37

The power lost $P_{lost}=P_{input}-P_{output}$. Given $P_{input}=500\ mW$ and $P_{output}=400\ mW$, so $P_{lost}=500 - 400=100\ mW$. The efficiency $\eta=\frac{P_{output}}{P_{input}}\times100%$. Substituting the values, $\eta=\frac{400}{500}\times 100% = 80%$.

Step6: Solve problem 38

Given $\eta = 85%=0.85$ and $P_{output}=5\ W$. Since $\eta=\frac{P_{output}}{P_{input}}$, we can find $P_{input}=\frac{P_{output}}{\eta}$. Substituting the values, $P_{input}=\frac{5}{0.85}\approx5.88\ W$.

Answer:

Problem 34: 36 Ah Problem 35: 8 A Problem 36: 13.5 mA Problem 37: Power lost: 100 mW, Efficiency: 80% Problem 38: 5.88 W