RC Circuit Transient Surge (Electrical Engineering) Solution

Problem Overview

We calculate the sensitivity of a discharging capacitor's voltage to ambient operational temperature changes ($\frac{dV}{d\theta}$) in an automated industrial circuit where clock speed drifts with temperature:

Step-by-Step Mathematical Solution

Step 1: Identify Composite Structure via Leibniz Notation

Because voltage $V$ depends on time $t$, and time parameter $t$ depends on temperature $\theta$, the Chain Rule in Leibniz notation is:

Step 2: Differentiate Component Functions Independently

1. Voltage with respect to time $\frac{dV}{dt}$: Using the exponential derivative rule $\frac{d}{dt}[e^u] = e^u u'$:
   $$\frac{dV}{dt} = 12 \cdot (-2)e^{-2t} = -24e^{-2t}\text{ V/s}$$
2. Time with respect to temperature $\frac{dt}{d\theta}$: Using the Power Rule term-by-term:
   $$\frac{dt}{d\theta} = \frac{d}{d\theta}[0.5\theta^2 + \theta] = 2(0.5)\theta + 1 = \theta + 1\text{ s/}^\circ\text{C}$$

Step 3: Combine Rates via the Chain Rule

Step 4: Evaluate Operating Conditions at $\theta = 2^\circ\text{C}$

1. Find elapsed time $t$ at $\theta = 2^\circ\text{C}$:
   $$t(2) = 0.5(2)^2 + 2 = 0.5(4) + 2 = 2 + 2 = 4\text{ seconds}$$
2. Substitute $\theta = 2$ and $t = 4$ into $\frac{dV}{d\theta}$:
   $$\frac{dV}{d\theta} = -24e^{-2(4)} \cdot (2 + 1) = -24e^{-8} \cdot 3 = -72e^{-8}\text{ V/}^\circ\text{C}$$
   Numerically:
   $$-72e^{-8} = -\frac{72}{e^8} \approx -0.02415\text{ V/}^\circ\text{C}$$

Engineering Conclusion

At an operating temperature of $\theta = 2^\circ\text{C}$, the capacitor voltage decreases at approximately $0.0242\text{ Volts per }^\circ\text{C}$ due to thermal timing drift. Electrical hardware engineers use this sensitivity metric to program thermal compensation look-up tables into microcontroller firmware, preventing false low-voltage triggers in safety-critical automated shutdown systems.