To find the terminal voltage of the generator, we need to follow a systematic approach.

Step 1: Analyze the given information

• The generator provides a 12 V output under ideal conditions (no load).
• The generator's internal resistance, $R_g = 0.05 \, \Omega$.
• Two loads are connected in parallel:
  1. The first load draws a current of 12 A.
  2. The second load dissipates power at a rate of 200 W.

Step 2: Calculate the current for the second load

The power dissipated by a load is related to its voltage and current through the formula: $P = V \times I$ Rearranging this to solve for the current $I_2$ drawn by the second load: $I_2 = \frac{P}{V}$ We don’t know the terminal voltage $V_t$ yet, but this is the same voltage across both loads. Let’s leave this as an equation for now: $I_2 = \frac{200}{V_t}$

Step 3: Total current supplied by the generator

The total current $I_{total}$ supplied by the generator is the sum of the currents for both loads, which gives: $I_{total} = I_1 + I_2 = 12 + \frac{200}{V_t}$

Step 4: Voltage drop due to internal resistance

The generator has an internal resistance, so the voltage at the generator’s terminals will drop by an amount $I_{total} \times R_g$, meaning the terminal voltage is: $V_t = 12 - I_{total} \times 0.05$ Substituting $I_{total}$ from Step 3 into this equation: $V_t = 12 - \left(12 + \frac{200}{V_t}\right) \times 0.05$ This is a nonlinear equation in $V_t$, and we can solve it to find the terminal voltage.

Step 5: Solve for $V_t$

Let's now solve this equation for $V_t$.

$V_t = 12 - \left(12 + \frac{200}{V_t}\right) \times 0.05$ $V_t = 12 - 0.05 \times 12 - 0.05 \times \frac{200}{V_t}$ $V_t = 12 - 0.6 - \frac{10}{V_t}$ $V_t + \frac{10}{V_t} = 11.4$

Multiplying both sides by $V_t$ to eliminate the fraction:

$V_t^2 + 10 = 11.4 V_t$

Rearranging this into a quadratic form:

$V_t^2 - 11.4 V_t + 10 = 0$

Step 6: Solve the quadratic equation

Now we solve this quadratic equation using the quadratic formula: $V_t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where $a = 1$, $b = -11.4$, and $c = 10$. Plugging these values into the formula:

$V_t = \frac{-(-11.4) \pm \sqrt{(-11.4)^2 - 4(1)(10)}}{2(1)}$ $V_t = \frac{11.4 \pm \sqrt{130.56 - 40}}{2}$ $V_t = \frac{11.4 \pm \sqrt{90.56}}{2}$ $V_t = \frac{11.4 \pm 9.51}{2}$

Now, solving for the two possible values of $V_t$:

1. $V_t = \frac{11.4 + 9.51}{2} = \frac{20.91}{2} = 10.455 \, \text{V}$
2. $V_t = \frac{11.4 - 9.51}{2} = \frac{1.89}{2} = 0.945 \, \text{V}$

Since the terminal voltage must be reasonably close to 12 V, the correct value is: $V_t = 10.455 \, \text{V}$

Conclusion:

The terminal voltage of the generator at this load is approximately 10.46 V.

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Do you want more details or have any questions?

Here are 5 related questions for further practice:

1. How would the terminal voltage change if the generator's internal resistance was higher?
2. What would happen to the terminal voltage if only the 12 A load was connected?
3. How would the power dissipation in the second load change if the terminal voltage increased?
4. What is the total power supplied by the generator at this load?
5. How does the efficiency of the generator change with the internal resistance?

Tip: In circuits with internal resistance, the total power is shared between the load and the internal losses. Always account for the voltage drop due to internal resistance to find accurate terminal voltage.