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 \((\mathrm{a})\) the charge on the capacitor, \((\mathrm{b})\) the current into the capacitor, and (c) the power output from the power supply.

Step-by-Step Solution

Answer

(a) 225 μC, (b) 60 μA, (c) 450 μW.

Step 1: Determine the voltage at t = 0.500 s

We start by finding the voltage across the capacitor at the given time, which is calculated using the given voltage function. Substitute \( t = 0.500 \) s into \( V(t) = 6.00 + 4.00t - 2.00t^2 \): \[ V(0.500) = 6.00 + 4.00(0.500) - 2.00(0.500)^2 \] Simplifying, we find:\[ V(0.500) = 6.00 + 2.00 - 0.500 = 7.50 \text{ volts} \]

Step 2: Calculate the charge on the capacitor

The charge \( Q \) on a capacitor is given by the formula \( Q = C \times V \), where \( C \) is the capacitance and \( V \) is the voltage. Using the capacitance \( C = 30 \mu F = 30 \times 10^{-6} \) F and the voltage \( V = 7.50 \) volts from Step 1, we calculate:\[ Q = 30 \times 10^{-6} \times 7.50 = 225 \times 10^{-6} \text{ coulombs} = 225 \mu C \]

Step 3: Determine the current into the capacitor

The current \( I \) into a capacitor is the derivative of charge with respect to time. Differentiate the voltage function to find \( \frac{dV}{dt} \):\[ \frac{dV}{dt} = \frac{d}{dt}(6.00 + 4.00t - 2.00t^2) = 4.00 - 4.00t \]At \( t = 0.500 \) s, substitute in:\[ \frac{dV}{dt}(0.500) = 4.00 - 4.00(0.500) = 2.00 \text{ V/s} \]Now, using \( I = C \times \frac{dV}{dt} \):\[ I = 30 \times 10^{-6} \times 2.00 = 60 \times 10^{-6} \text{ A} = 60 \mu A \]

Step 4: Calculate the power output from the power supply

The power \( P \) can be calculated using the formula \( P = V \times I \). Using \( V = 7.50 \) volts and \( I = 60 \times 10^{-6} \) A from previous steps:\[ P = 7.50 \times 60 \times 10^{-6} = 450 \times 10^{-6} \text{ watts} = 450 \mu W \]

Key Concepts

Capacitor Charge

Capacitors are essential components in electrical circuits, as they store electrical charge. The charge stored in a capacitor, denoted by \( Q \), is determined by its capacitance \( C \) and the voltage \( V \) across it, following the formula \( Q = C \times V \). In the given problem, we are dealing with a \( 30 \mu F \) capacitor. This is a typical capacitance value, where \( \mu F \) denotes microfarads, a unit for measuring capacitance.
For this specific capacitor, the charge stored depends on the voltage at a specific time. By substituting the time \( t = 0.500 \) seconds into the voltage function \( V(t) = 6.00 + 4.00t - 2.00t^2 \), we determine that the voltage at this instant is \( 7.50 \) volts. Consequently, using the formula for capacitor charge, \( Q = 30 \times 10^{-6} \times 7.50 \), we find that the capacitor holds a charge of \( 225 \mu C \) (microcoulombs).

• Capacitance \( C \): Defines how much charge a capacitor can store per volt.
• Voltage \( V \): Determines how much charge is pushed into the capacitor at any given time.
• Total Stored Charge \( Q \): The total charge held at that voltage.

Electric Current

Electric current represents the flow of electric charge in a circuit and is crucial in how circuits operate. In a capacitor circuit, the current into the capacitor can be determined by observing how quickly its voltage changes with time. In mathematical terms, current \( I \) is calculated as the product of capacitance \( C \) and the rate of change of voltage over time, given by \( \frac{dV}{dt} \).
For our problem, this translates to differentiating the voltage function \( V(t) = 6.00 + 4.00t - 2.00t^2 \), resulting in \( \frac{dV}{dt} = 4.00 - 4.00t \). At \( t = 0.500 \), substituting this value gives \( \frac{dV}{dt} = 2.00 \) V/s. The current can now be calculated as \( I = 30 \times 10^{-6} \times 2.00 = 60 \mu A \) (microamperes).

• Current \( I \): Indicates the flow of charge per unit of time.
• Rate of Change of Voltage \( \frac{dV}{dt} \): Reflects how voltage varies over time, which impacts current.
• Overall, current relates both to the stored energy and the rate of charging or discharging in capacitors.

Electric Power

Electric power in a circuit refers to the rate at which energy is used or transferred. It is a vital aspect, determining how electrical devices and components function efficiently and safely. In our context, the power output of the power supply can be calculated using the simple formula \( P = V \times I \), where \( P \) represents power, \( V \) is the voltage, and \( I \) is the current.
From the previous computations, we have obtained the voltage across the capacitor \( V = 7.50 \) volts, and the current entering it \( I = 60 \times 10^{-6} \) amperes. Plugging these values into the formula gives us \( P = 7.50 \times 60 \times 10^{-6} = 450 \mu W \) (microwatts).

• Power \( P \): Measures energy transfer or conversion per unit of time.
• This formula highlights how both voltage and current influence power output.
• Understanding power is crucial for evaluating performance and energy efficiency.

Voltage Function

A voltage function describes how voltage varies with time in an electrical circuit. It can be linear, exponential, or even quadratic, as seen in this exercise's function, \( V(t) = 6.00 + 4.00t - 2.00t^2 \). This quadratic equation varies with time due to changing coefficients, reflecting more complex behavior typical in programmed power supplies or dynamic systems.
The voltage function directly impacts how much charge can accumulate on a capacitor and how the current behaves. By evaluating the voltage function at specific times, such as \( t = 0.500 \) seconds, we get real-time insights into circuit behavior. This specific analysis showed a voltage increment up to \( 7.50 \) volts during the particular interval.

• Voltage functions are pivotal in predicting and understanding circuit dynamics.
• They help in calculating other aspects like charge and current at any given instant.
• With dynamic voltage, the system adjusts its behavior over time accordingly.