Problem 85 - Fundamentals of Physics

Question

A \(30 \mu \mathrm{F}\) capacitor is connected across a programmed power supply. During the interval from \(t=0\) to \(t=3.00 \mathrm{~s}\) the output voltage of the supply is given by \(V(t)=6.00+4.00 t-2.00 t^{2}\) volts. At \(t=0.500 \mathrm{~s}\) find (a) the charge on the capacitor, (b) the current into the capacitor, and (c) the power output from the power supply.

Step-by-Step Solution

Answer

(a) \(225 \mu \text{C}\), (b) \(60 \mu \text{A}\), (c) \(450 \mu \text{W}\).

Step 1: Understand the Problem

You have a capacitor and a varying voltage over time. You need to find the charge on the capacitor, the current into it, and the power output from the power supply at a specific time, which is at \( t=0.5\, \text{s} \).

Step 2: Find the Charge on the Capacitor

The charge \( Q \) on a capacitor is given by the formula \( Q = C \times V(t) \), where \( C \) is the capacitance and \( V(t) \) is the voltage at time \( t \). Here, \( C = 30 \times 10^{-6} \text{ F} \) and \( V(t) = 6.00 + 4.00t - 2.00t^2 \). Substitute \( t = 0.5 \) into \( V(t) \) to find \( V(0.5) = 6.00 + 4.00 \times 0.5 - 2.00 \times (0.5)^2 = 7.5 \text{ V} \). Then calculate the charge: \( Q = 30 \times 10^{-6} \text{ F} \times 7.5 \text{ V} = 225 \times 10^{-6} \text{ C} = 225 \mu \text{C} \).

Step 3: Find the Current into the Capacitor

The current \( I \) into a capacitor is given by the derivative of the charge with respect to time, \( I = \frac{dQ}{dt} = C \cdot \frac{dV}{dt} \). First, find \( \frac{dV}{dt} \) by differentiating \( V(t) = 6.00 + 4.00t - 2.00t^2 \): \( \frac{dV}{dt} = 4.00 - 4.00t \). At \( t = 0.5 \), \( \frac{dV}{dt} = 4.00 - 4.00 \times 0.5 = 2.00 \). Thus, \( I = 30 \times 10^{-6} \times 2.00 = 60 \times 10^{-6} \text{ A} = 60 \mu \text{A} \).

Step 4: Find the Power Output from the Power Supply

Power \( P \) is calculated as \( P = I \times V \). From the previous steps, \( I = 60 \mu \text{A} \) and \( V = 7.5 \text{ V} \). Therefore, \( P = 60 \times 10^{-6} \text{ A} \times 7.5 \text{ V} = 450 \times 10^{-6} \text{ W} = 450 \mu \text{W} \).

Key Concepts

Charge on Capacitor

The charge on a capacitor is fundamental to understanding its function in an electric circuit. Capacitors store and release electrical energy, which is quantified by the amount of charge they hold. To determine the charge, we use the formula:

• \( Q = C \times V(t) \)

where \( Q \) is the charge, \( C \) is the capacitance, and \( V(t) \) is the voltage at time \( t \). In our exercise, the capacitance \( C \) is \( 30 \mu \text{F} \). The voltage at \( t = 0.5 \, \text{s} \) is calculated by substituting into the equation \( V(t) = 6.00 + 4.00t - 2.00t^2 \).
This gives us \( V(0.5) = 7.5 \, \text{V} \). Thus, the charge at this moment is

• \( Q = 30 \times 10^{-6} \text{ F} \times 7.5 \text{ V} = 225 \mu \text{C} \).

Understanding capacitor charge is crucial in circuits as it affects how energy is stored and released.
It also influences how quickly a capacitor can respond to changes in voltage.

Current into Capacitor

The current flowing into a capacitor changes over time as it fills with charge. This current is closely linked with the rate of change of voltage across the capacitor. To find the current, we use the relationship:

• \( I = \frac{dQ}{dt} = C \cdot \frac{dV}{dt} \)

For this, we need to determine the derivative of the voltage function \( V(t) \). By differentiating \( V(t) = 6.00 + 4.00t - 2.00t^2 \), we obtain:

• \( \frac{dV}{dt} = 4.00 - 4.00t \)

At \( t = 0.5 \, \text{s} \), the rate of change of voltage \( \frac{dV}{dt} \) is \( 2.00 \). Therefore, the current \( I \) is:

• \( I = 30 \times 10^{-6} \times 2.00 = 60 \mu \text{A} \).

Current into a capacitor is a key metric as it influences charging time. It's important in applications where timing is crucial, such as in oscillators and timing circuits.

Power Output

Power output from a power supply to a capacitor is essential for understanding the energy dynamics in circuits. Power indicates how much energy per unit time is being transferred to the capacitor. It's calculated as:

• \( P = I \times V \)

where \( I \) is the current and \( V \) is the voltage. From previous calculations, we have \( I = 60 \mu \text{A} \) and \( V = 7.5 \text{ V} \). Thus, the power output is:

• \( P = 60 \times 10^{-6} \text{ A} \times 7.5 \text{ V} = 450 \mu \text{W} \).

Understanding power output is vital for designing efficient electronic circuits. Inefficient management of power can lead to unnecessary energy loss and overheating, impacting performance.