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We are given the following information:

1. Ohm's law: \( V = IR \), where:
   - \( V \) is the voltage,
   - \( I \) is the current,
   - \( R \) is the resistance.

2. For two circuits with resistances \( R_1 \) and \( R_2 \) in parallel, the combined resistance \( R \) is given by:
   \[
   \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}
   \]

We are asked to calculate the rate of change of voltage (\( \frac{dV}{dt} \)) when:
- \( I = 2 \) amps and \( \frac{dI}{dt} = 0.02 \) amps/sec,
- \( R_1 = 3 \) ohms and \( \frac{dR_1}{dt} = 0.5 \) ohms/sec,
- \( R_2 = 4 \) ohms and \( \frac{dR_2}{dt} = -0.2 \) ohms/sec.

### Step 1: Find the combined resistance \( R \)
Using the formula for combined resistance:
\[
\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}
\]
Substitute the given values of \( R_1 = 3 \) and \( R_2 = 4 \):
\[
\frac{1}{R} = \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}
\]
Thus,
\[
R = \frac{12}{7} \approx 1.714 \text{ ohms}.
\]

### Step 2: Differentiate Ohm's Law
To find the rate of change of voltage, we differentiate \( V = IR \) with respect to time:
\[
\frac{dV}{dt} = I \frac{dR}{dt} + R \frac{dI}{dt}.
\]
We need \( \frac{dR}{dt} \), which we can find by differentiating the combined resistance formula.

### Step 3: Differentiate the combined resistance formula
Differentiate both sides of \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \) with respect to time:
\[
-\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt}.
\]
Substitute the known values of \( R_1 \), \( R_2 \), \( \frac{dR_1}{dt} \), and \( \frac{dR_2}{dt} \):
\[
-\frac{1}{(1.714)^2} \frac{dR}{dt} = -\frac{1}{(3)^2}(0.5) - \frac{1}{(4)^2}(-0.2).
\]
Simplifying:
\[
-\frac{1}{2.94} \frac{dR}{dt} = -\frac{0.5}{9} + \frac{0.2}{16}.
\]
Calculate each term:
\[
-\frac{1}{2.94} \frac{dR}{dt} = -0.0556 + 0.0125 = -0.0431.
\]
Now solve for \( \frac{dR}{dt} \):
\[
\frac{dR}{dt} = (-0.0431)(-2.94) \approx 0.1267 \, \text{ohms/sec}.
\]

### Step 4: Calculate \( \frac{dV}{dt} \)
Now, substitute the known values into the differentiated form of Ohm’s law:
\[
\frac{dV}{dt} = (2)(0.1267) + (1.714)(0.02).
\]
Simplify:
\[
\frac{dV}{dt} = 0.2534 + 0.03428 \approx 0.2877 \, \text{volts/sec}.
\]

Thus, the rate at which the voltage is changing is approximately \( 0.288 \, \text{volts/sec} \).

Would you like any further details or have any questions?

### Related Questions:
1. How does the rate of change of resistance affect voltage over time?
2. Can you explain how resistors in series would change this problem's approach?
3. What happens to voltage if the current is decreasing instead of increasing?
4. How would this problem change if \( R_1 \) and \( R_2 \) were in series?
5. Can this problem be solved graphically, and how would the curves look?

#### Tip:
When working with multiple variables changing over time, always identify which values you need to differentiate implicitly to understand their relationship.

The voltage, V (in volts), across a circuit is given by Ohm's law: V = IR, where I is the current (in amps) flowing through the circuit and R is the resistance (in ohms). If we place two circuits, with resistance R1 and R2, in parallel, then their combined resistance, R, is given by 1/R = 1/R1 + 1/R2. Suppose the current is 2 amps and increasing at 0.02 amps/sec and R1 is 3 ohms and increasing at 0.5 ohms/sec, while R2 is 4 ohms and decreasing at 0.2 ohms/sec. Calculate the rate at which the voltage is changing.

This problem explores how to calculate the rate of voltage change using Ohm's law and the formula for combined resistance of two parallel circuits. It involves differentiating key electrical formulas and applying implicit differentiation. The solution shows step-by-step calculations with values of current and resistance changing over time. Suitable for Grades 11-12.

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