To find the rate at which the voltage is changing, we need to consider how voltage depends on current and resistance, as well as how the resistances $R_1$ and $R_2$ change with time.

Given:

• Current $I = 4$ amps and increasing at a rate of $\frac{dI}{dt} = 0.01$ amps/sec
• Resistance $R_1 = 5$ ohms and increasing at a rate of $\frac{dR_1}{dt} = 0.5$ ohms/sec
• Resistance $R_2 = 3$ ohms and decreasing at a rate of $\frac{dR_2}{dt} = -0.3$ ohms/sec

Ohm's Law is given by: $V = I \cdot R$ where $R$ is the total resistance in the parallel circuit, and $I$ is the current.

Step 1: Total Resistance in Parallel

The total resistance $R$ for two resistors $R_1$ and $R_2$ in parallel is given by: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ Solving for $R$, we get: $R = \frac{R_1 R_2}{R_1 + R_2}$

Step 2: Differentiate the Voltage

We need to calculate the rate of change of voltage with respect to time, $\frac{dV}{dt}$.

Using the product rule for differentiation: $V = I \cdot R$ Differentiate both sides with respect to time $t$: $\frac{dV}{dt} = \frac{d}{dt}(I \cdot R) = \frac{dI}{dt} \cdot R + I \cdot \frac{dR}{dt}$

Step 3: Calculate $\frac{dR}{dt}$

We need to differentiate $R = \frac{R_1 R_2}{R_1 + R_2}$ with respect to time $t$, using the quotient rule: $\frac{dR}{dt} = \frac{(R_1 + R_2) \cdot \frac{d}{dt}(R_1 R_2) - (R_1 R_2) \cdot \frac{d}{dt}(R_1 + R_2)}{(R_1 + R_2)^2}$ Expanding these derivatives:

$\frac{d}{dt}(R_1 R_2) = \frac{dR_1}{dt} \cdot R_2 + R_1 \cdot \frac{dR_2}{dt}$ $\frac{d}{dt}(R_1 + R_2) = \frac{dR_1}{dt} + \frac{dR_2}{dt}$

Now plug in the values:

• $\frac{dR_1}{dt} = 0.5$ ohms/sec
• $\frac{dR_2}{dt} = -0.3$ ohms/sec
• $R_1 = 5$ ohms
• $R_2 = 3$ ohms

Let’s calculate $\frac{dR}{dt}$:

$\frac{dR}{dt} = \frac{(5 + 3) \cdot (0.5 \cdot 3 + 5 \cdot -0.3) - (5 \cdot 3) \cdot (0.5 + -0.3)}{(5 + 3)^2}$ Now, solving the equation step by step:

$\frac{dR}{dt} = \frac{8 \cdot (1.5 - 1.5) - 15 \cdot (0.5 - 0.3)}{64}$ $\frac{dR}{dt} = \frac{8 \cdot 0 - 15 \cdot 0.2}{64}$ $\frac{dR}{dt} = \frac{-3}{64} \quad \text{ohms/sec}$

Step 4: Calculate $\frac{dV}{dt}$

Now we have everything we need to calculate the rate of change of the voltage:

$\frac{dV}{dt} = \frac{dI}{dt} \cdot R + I \cdot \frac{dR}{dt}$

Substitute the values:

• $\frac{dI}{dt} = 0.01$ amps/sec
• $I = 4$ amps
• $R = \frac{5 \cdot 3}{5 + 3} = \frac{15}{8} = 1.875$ ohms
• $\frac{dR}{dt} = \frac{-3}{64}$ ohms/sec

Now substitute into the formula: $\frac{dV}{dt} = 0.01 \cdot 1.875 + 4 \cdot \left(\frac{-3}{64}\right)$ $\frac{dV}{dt} = 0.01875 - 0.1875$ $\frac{dV}{dt} = -0.16875 \quad \text{volts/sec}$

Final Answer:

The voltage is decreasing at a rate of $0.16875$ volts per second.

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Would you like more details on any part of the calculation? Here are some related questions:

1. How would the calculation change if the resistors were in series instead of parallel?
2. What is the effect on voltage if the current increases at a faster rate?
3. How do the rates of change of current and resistance affect the voltage?
4. What would happen to the rate of change of voltage if both resistances increased instead of one increasing and the other decreasing?
5. How can Ohm’s Law be applied in circuits with more than two resistors in parallel?

Tip: Always ensure the units are consistent (amps for current, ohms for resistance) when performing calculations with rates of change!