Question

problem 32?

1. a battery can provide an average of 1.5 a of current for 24 h. what is its ampere - hour rating?
2. how much average current can be drawn from an 80 ah battery for 10 h?
3. if a battery is rated at 650 mah, how much average current will it provide for 48 h?
4. if the input power is 500 mw and the output power is 400 mw, how much power is lost? what is the efficiency of this power supply?
5. to operate at 85% efficiency, how much output power must a source produce if the input power is 5 w?

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