A battery discharge follows the formula v = V0 * e^(-t/RC). Given V0 = 12V, R = 0.50 Ohms, C = 100F, determine the time t when the voltage drops to 11.0V and transpose the formula to make R the subjec

The formula for battery discharge is:

$$v = V_0 e^{\left(-\frac{t}{RC}\right)}$$

Given values are $V_0 = 12V$, $R = 0.50 \, \Omega$, $C = 100F$. We want to find $t$ when $v = 11.0V$.

$$11.0V = 12V \cdot e^{\left(-\frac{t}{0.50 \, \Omega \times 100F}\right)}$$ $$11.0 = 12 \cdot e^{\left(-\frac{t}{50}\right)}$$

$$\ln\left(\frac{11.0}{12}\right) = -\frac{t}{50}$$ $$t = -50 \cdot \ln\left(\frac{11.0}{12}\right)$$ $$t = -50 \cdot (-0.08701)$$ $$t = 4.3505 \, \text{seconds}$$

The time when the battery voltage drops to $11.0V$ is $\boxed{\text{4.35 s}}$.

A.2: Transpose the formula to make $R$ the subject.

$$R = \frac{-t}{C \ln\left(\frac{v}{V_0}\right)}$$

The formula transposed for $R$ is $\boxed{R = \frac{-t}{C \ln\left(\frac{v}{V_0}\right)}}$.

Task 1, Part B:

B.i: Determine the power will be supplied by a deep-cycle battery.

Given voltage $V = 12V$ and current $I = 5A$.

$$P = 12V \times 5A$$ $$P = 60W$$

The power supplied by the battery is $\boxed{\text{60 W}}$.

B.ii: Determine the total electrical energy (in Watt-hours) consumed by the pump during its daily 2-hour operation.

From B.i, power $P = 60W$. The pump operates for $t = 2$ hours.

$$E = 60W \times 2 \, \text{hours}$$ $$E = 120 \, \text{Wh}$$

The total electrical energy consumed is $\boxed{\text{120 Wh}}$.

B.iii: Calculate the percentage of its stored energy will be used by the pump's daily operation.

Energy consumed $E_{consumed} = 120 \, \text{Wh}$ (from B.ii).
Total usable energy $E_{total} = 1200 \, \text{Wh}$.

$$\text{Percentage used} = \left(\frac{E_{consumed}}{E_{total}}\right) \times 100\%$$ $$\text{Percentage used} = \left(\frac{120 \, \text{Wh}}{1200 \, \text{Wh}}\right) \times 100\%$$ $$\text{Percentage used} = 0.1 \times 100\%$$ $$\text{Percentage used} = 10\%$$

The percentage of stored energy used is \boxed{\text{10%}}.

B.iv: Calculate the power dissipated as heat inside the battery while it is supplying the 5A current to the pump.

Given current $I = 5A$ and internal resistance $R_{internal} = 0.1 \, \Omega$.

$$P_{dissipated} = (5A)^2 \times 0.1 \, \Omega$$ $$P_{dissipated} = 25A^2 \times 0.1 \, \Omega$$ $$P_{dissipated} = 2.5W$$

The power dissipated as heat is $\boxed{\text{2.5 W}}$.

B.v: Calculate the total charge (in Coulombs) delivered to the battery during this time.

Given charging current $I = 4A$ and time $t = 5$ hours.
Convert time from hours to seconds:

$$t = 5 \, \text{hours} \times 3600 \, \text{seconds/hour} = 18000 \, \text{seconds}$$

$$Q = 4A \times 18000 \, \text{s}$$ $$Q = 72000 \, \text{C}$$

The total charge delivered to the battery is $\boxed{\text{72000 C}}$.

B.vi: Draw a circuit diagram showing how to connect the four panels to achieve 48V output.

To achieve a 48V output from four solar panels, assuming each panel has a nominal voltage of 12V (a common rating for solar panels in 12V systems), the panels must be connected in series.

A circuit diagram would show:

• Four solar panel symbols connected in a chain.
• The positive terminal of the first panel connects to the negative terminal of the second panel.
• The positive terminal of the second panel connects to the negative terminal of the third panel.
• The positive terminal of the third panel connects to the negative terminal of the fourth panel.
• The overall output would be taken from the negative terminal of the first panel and the positive terminal of the fourth panel. This series connection adds the voltages of individual panels: $12V + 12V + 12V + 12V = 48V$.