Problem 25 - Basic Technical Mathematics with Calculus

Question

The voltage across a \(2.5-\mu\) F capacitor in a copying machine is zero. What is the voltage after 12 ms if a current of 25 mA charges the capacitor?

Step-by-Step Solution

Answer

The voltage across the capacitor after 12 ms is approximately 120 volts.

Step 1: Understanding the Capacitor Charging Formula

When a capacitor charges, the relationship between the charge, capacitance, and voltage is given by the formula: \[ q = C imes V \] where \( q \) is the charge, \( C \) is the capacitance, and \( V \) is the voltage. Initially, the voltage across the capacitor is zero.

Step 2: Calculate the Charge After Time

The change in charge \( \Delta q \) on the capacitor can also be expressed in terms of current \( I \) over time \( t \) as:\[ \Delta q = I imes t \]Given a current \( I = 25 \) mA (or \( 0.025 \) A) and time \( t = 12 \) ms (or \( 0.012 \) s), we find:\[ \Delta q = 0.025 \times 0.012 = 0.0003 \] coulombs.

Step 3: Relate Charge to Voltage

Revisit the formula for capacitance and voltage:\[ q = C imes V \]Rearrange it to solve for voltage \( V \):\[ V = \frac{q}{C} \]Now substitute the values \( \Delta q = 0.0003 \) C and \( C = 2.5 \times 10^{-6} \) F:\[ V = \frac{0.0003}{2.5 \times 10^{-6}} \approx 120 \] volts.

Key Concepts

Capacitance and Voltage

When discussing capacitors in electronics, it's essential to understand the relationship between capacitance and voltage.
Capacitance, denoted as \( C \), measures a capacitor's ability to store an electric charge per unit voltage. The unit of capacitance is the farad (F). In our example, the capacitor in the copying machine has a capacitance of \( 2.5 \underline{\phantom{xxx}}\mu\text{F}\).

The voltage \( V \) across the capacitor describes the electric potential difference between its plates. As a capacitor charges, the voltage increases until it reaches the voltage of the power source.

• The relationship between these quantities is given by the formula \( q = C \times V \), where \( q \) is the charge stored.
• This means, for a given amount of stored charge, the voltage is directly proportional to capacitance.
• Upon charging, the initially zero voltage across the capacitor will rise as charge accumulates across its plates.

This interplay is what governs how capacitors behave in circuits, as they temporarily store and release energy.

Current and Charge Relationship

Understanding how current and charge relate is crucial for analyzing capacitor charging. Current \( I \), measured in amperes (A), is the flow of electrical charge over time. For our problem, the current charging the capacitor is 25 mA, which is equivalent to 0.025 A.

When a constant current flows into a capacitor, it accumulates charge. The relationship between current \( I \), charge \( \Delta q \), and time \( t \) is expressed as:

• \( \Delta q = I \times t \)

When we apply a current of 0.025 A for 12 milliseconds (or 0.012 seconds), we calculate the charge added to the capacitor as:
\[ \Delta q = 0.025 \times 0.012 = 0.0003 \]coulombs.

This formula highlights how increasing either the current or the duration of the current flow directly increases the amount of charge stored on the capacitor. More charge means a higher voltage across the capacitor due to its capacitance characteristic.

Charge Calculation

The charge stored in a capacitor can be used to determine the voltage across it. Let's break down how this calculation works using our example:

Firstly, we know the charge \( \Delta q \) that has built up during a given time as current flows. This charge is calculated to be 0.0003 coulombs in our case.

The relationship between charge, capacitance \( C \), and voltage \( V \) is straightforward:

• From \( q = C \times V \), we rearrange it to find voltage: \( V = \frac{q}{C} \).

Substituting our known values into this equation allows us to find the voltage across the capacitor:
\[ V = \frac{0.0003}{2.5 \times 10^{-6}} \approx 120 \text{ volts} \]

This relationship is at the heart of capacitor functionality—storing charge which translates into voltage. Solving such equations helps predict how the capacitor behaves as the charge varies. Understanding these principles is crucial for anyone looking to work with or design electronic circuits.