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To solve part **(d)** of the problem, where we need to find the rate of change of resistance \( \frac{dR}{dt} \), we can use the equation given in part **(c)**:

\[
\frac{dR}{dt} = \frac{1}{I} \left( \frac{dV}{dt} - V \frac{dI}{dt} \right)
\]

### Given Data:
- \( V = 16 \text{ volts} \)
- \( I = 4 \text{ amperes} \)
- \( \frac{dV}{dt} = 5 \text{ volts/sec} \)
- \( \frac{dI}{dt} = -\frac{1}{2} \text{ amp/sec} \) (since the current is decreasing)

### Step-by-Step Calculation:
1. Substitute the known values into the formula:
\[
\frac{dR}{dt} = \frac{1}{4} \left( 5 - 16 \times \left(-\frac{1}{2}\right) \right)
\]
2. Simplify the expression:
\[
\frac{dR}{dt} = \frac{1}{4} \left( 5 + 8 \right) = \frac{1}{4} \times 13 = 3.25 \text{ ohms/sec}
\]

Thus, the resistance \( R \) is increasing at a rate of **3.25 ohms/sec**.

Would you like further details or have any other questions?

### 5 Related Questions:
1. How does Ohm's Law \( V = IR \) relate voltage, current, and resistance in this problem?
2. What does it mean physically when \( \frac{dR}{dt} \) is positive or negative?
3. How would the rate of change of \( R \) be affected if the current was increasing instead of decreasing?
4. Can you derive \( \frac{dR}{dt} \) directly from the product rule applied to \( V = IR \)?
5. How would changes in both voltage and current affect the resistance in a real electrical circuit?

### Tip:
Always double-check units when solving physics problems to ensure that your final answer has the correct dimensions.

The voltage V (volts), current I (amperes), and resistance R (ohms) of an electric circuit are related by the equation V = IR. Suppose V is increasing at the rate of 5 volts/sec while I is decreasing at the rate of -1/2 amp/sec. Find the rate at which R is changing when V = 16 volts and I = 4 amps.

Learn how to calculate the rate of change of resistance in an electrical circuit, where V = IR. Given that voltage is increasing at 5 volts/sec and current is decreasing at -1/2 amp/sec, find dR/dt when V = 16 volts and I = 4 amps. This problem involves differentiation and Ohm’s Law.

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