Problem 22 - Basic Technical Mathematics with Calculus

Question

The voltage across a \(0.10-\mu \mathrm{F}\) capacitor in a microwave oven is zero. What is the voltage after being charged by a \(0.25 \mu\) A current for \(3.0 \mu \mathrm{s} ?\)

Step-by-Step Solution

Answer

The voltage across the capacitor is now \(7.5 \times 10^{-6}\) volts.

Step 1: Understanding the relationship

The relationship between current \(I\), capacitance \(C\), change in voltage \(\Delta V\), and time \(t\) for charging a capacitor is given by the formula: \(\Delta V = \frac{I \, t}{C}\). We will use this formula to find the change in voltage across the capacitor.

Step 2: Identifying known values

We know the following values: current \(I = 0.25 \, \mu A = 0.25 \times 10^{-6} \, A\), time \(t = 3.0 \, \mu s = 3.0 \times 10^{-6} \, s\), and capacitance \(C = 0.10 \, \mu F = 0.10 \times 10^{-6} \, F\).

Step 3: Calculating the change in voltage

Substitute the known values into the formula:\[\Delta V = \frac{(0.25 \times 10^{-6}) \, (3.0 \times 10^{-6})}{0.10 \times 10^{-6}}\]Simplifying the equation, we get:\[\Delta V = \frac{0.75 \times 10^{-12}}{0.10 \times 10^{-6}} = 7.5 \times 10^{-6} \, V\]This means the change in voltage across the capacitor is \(7.5 \times 10^{-6}\) volts.

Key Concepts

Current-Capacitance Relationship

When charging a capacitor, understanding the relationship between current, capacitance, and voltage is crucial. This relationship is described by the formula \( \Delta V = \frac{I \, t}{C} \). This tells us that the change in voltage (\(\Delta V\)) across a capacitor depends on the current (\(I\)), the time (\(t\)) during which the current flows, and the capacitance of the capacitor (\(C\)).

• Current is like the flow of water through a hose, indicating how much charge passes through the circuit per second.
• Capacitance measures a capacitor's ability to store charge and is determined by its physical characteristics, such as surface area and the distance between its plates.
• A higher capacitance means a more significant ability to store charge, resulting in a smaller voltage change for a given current and time.
• Conversely, a smaller capacitance means even small currents or short periods can lead to considerable voltage changes.

Understanding this equation helps us to determine how voltage develops across a capacitor during charging.

Voltage Calculation

Calculating voltage across a capacitor involves substituting the known values into the equation derived from the current-capacitance relationship. For our exercise, we're given a current of \(0.25 \, \mu A\) and a time interval of \(3.0 \, \mu s\). The capacitance is \(0.10 \, \mu F\).To find the change in voltage:

• First, convert all quantities to standard units:
  • Current \(I = 0.25 \, \mu A \rightarrow 0.25 \times 10^{-6} \) A
  • Time \(t = 3.0 \, \mu s \rightarrow 3.0 \times 10^{-6} \) s
  • Capacitance \(C = 0.10 \, \mu F \rightarrow 0.10 \times 10^{-6} \) F
• Substitute into the formula: \( \Delta V = \frac{(0.25 \times 10^{-6}) \, (3.0 \times 10^{-6})}{0.10 \times 10^{-6}} \)
• Simplify to get \( \Delta V = 7.5 \times 10^{-6} \, V \).

This process shows how small currents over short periods can result in measurable voltage changes across capacitors.

Charging Time

Charging time is a key factor in determining how much charge a capacitor stores, which directly affects the voltage across it. The longer a current flows into a capacitor, the greater the charge it accumulates and the higher the voltage it develops. Some important notes:

• Shorter time intervals mean less charge is accumulated if all other factors remain constant.
• For the same current and capacitance, extended charging time results in greater voltage across the capacitor.
• In our example, a charging time of \(3.0 \, \mu s\) was enough to produce a voltage change.
• Real-world applications may have different tolerances and requirements for charging time depending on the specifications of the device using the capacitor.

Understanding charging time helps in designing circuits where precise control over voltage is necessary.