Question

1. how much average current can be drawn from an 80 ah battery for 10 h?
2. if a battery is rated at 650 mah, how much average current will it provide for 48 h?
3. 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?
4. to operate at 85% efficiency, how much output power must a source produce if the input power is 5 w?

Explanation:

Step1: Recall the formula for current

The formula for current $I$ is $I=\frac{Q}{t}$, where $Q$ is the charge (in Ah - ampere - hour) and $t$ is the time in hours.

Step2: Solve problem 35

Given $Q = 80\ Ah$ and $t=10\ h$. Using the formula $I=\frac{Q}{t}$, we substitute the values: $I=\frac{80\ Ah}{10\ h}=8\ A$.

Step3: Solve problem 36

First, convert the charge from mAh to Ah. Given $Q = 650\ mAh=0.65\ Ah$ and $t = 48\ h$. Then, using $I=\frac{Q}{t}$, we have $I=\frac{0.65\ Ah}{48\ h}\approx0.0135\ A = 13.5\ mA$.

Step4: Solve problem 37

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

Step5: Solve problem 38

We know that $\eta=\frac{P_{out}}{P_{in}}\times100\%$. Given $\eta = 85\%=0.85$ and $P_{in}=5\ W$. Rearranging the formula for $P_{out}$, we get $P_{out}=\eta\times P_{in}=0.85\times5\ W = 4.25\ W$.

Answer:

1. $8\ A$
2. $13.5\ mA$
3. Power lost: $100\ mW$, Efficiency: $80\%$
4. $4.25\ W$