Question

A circuit has in series a constant electromotive force of \(100 \mathrm{~V}\), a resistor of \(10 \Omega\), and a capacitor of \(2 \times 10^{-4}\) farads. The switch is closed at time \(t=0\), and the charge on the capacitor at this instant is zero. Find the charge and the current at time \(t>0\)

Short answer

In conclusion, the current function is given by \(I(t) = 10 - 10e^{-\frac{t}{10(2 \times 10^{-4})}}\) A and the charge function is given by \(q(t) = 2 \times 10^{-2}(1 - e^{-\frac{t}{10(2 \times 10^{-4})}})\) C for any \(t > 0\).

Step-by-step solution

Determine voltage function across capacitor

The voltage across a capacitor is given by the equation: \(V_C(t) = V_0(1 - e^{-\frac{t}{RC}})\) In this problem, the electromotive force (battery voltage) is given as \(V_0 = 100\,\mathrm{V}\), the resistance is \(R = 10\, \Omega\), and the capacitance is \(C = 2 \times 10^{-4}\,\mathrm{F}\). We will plug these values into the equation to find the voltage across the capacitor as a function of time: \(V_C(t) = 100(1 - e^{-\frac{t}{10(2 \times 10^{-4})}})\)

Determine voltage function across resistor

In a series circuit, the sum of the voltages across all components is equal to the total voltage supplied by the EMF. Therefore, the voltage across the resistor at any given time can be found as: \(V_R(t) = V_0 - V_C(t)\) From step 1, we found the voltage across the capacitor as a function of time, so now we can plug it into the above equation: \(V_R(t) = 100 - 100(1 - e^{-\frac{t}{10(2 \times 10^{-4})}})\)

Determine current function in the circuit

Using Ohm's Law, we can determine the current function in the circuit: \(I(t) = \frac{V_R(t)}{R}\) Plugging in the voltage across the resistor and the resistance values, we get: \(I(t) = \frac{100 - 100(1 - e^{-\frac{t}{10(2 \times 10^{-4})}})}{10}\) Now we can simplify this expression: \(I(t) = 10 - 10e^{-\frac{t}{10(2 \times 10^{-4})}}\)

Determine charge function on the capacitor

We know that the charge on a capacitor is given by the equation: \(q(t) = CV_C(t)\) So, plugging in the values for the capacitance and the voltage function that we found earlier, we can determine the charge function: \(q(t) = 2 \times 10^{-4} \cdot 100(1 - e^{-\frac{t}{10(2 \times 10^{-4})}})\) Which simplifies to: \(q(t) = 2 \times 10^{-2}(1 - e^{-\frac{t}{10(2 \times 10^{-4})}})\) In conclusion, the current function is given by \(I(t) = 10 - 10e^{-\frac{t}{10(2 \times 10^{-4})}}\) A and the charge function is given by \(q(t) = 2 \times 10^{-2}(1 - e^{-\frac{t}{10(2 \times 10^{-4})}})\) C for any \(t > 0\).

Key concepts

Electromotive Force

Electric circuits often have a device that provides energy in the form of electricity. This energy source is usually referred to as electromotive force (EMF). EMF is not really a "force" in the traditional sense but rather a voltage, supplied by a battery or other energy source. It is the factor that drives the current through the circuit.
The standard unit for measuring EMF is volts (V), and it represents the potential difference between two points in an electric circuit. How does it work? Imagine a pump pushing water through pipes; similarly, the EMF causes electric charges to flow, creating an electric current.
In the exercise, the EMF is given as 100 V. This value indicates the energy provided to move the charges through the resistor and capacitor in the circuit.

Capacitors

Capacitors are fundamental components in many electrical circuits. They store electrical energy temporarily as an electric field. Think of them as devices that hold electric charge, similar to how a balloon holds air. When connected to a circuit, they can release this energy when needed.
The capacitance, measured in farads (F), indicates the amount of charge a capacitor can store per unit voltage across its plates. The higher the capacitance, the more charge it can store. In our exercise, the given capacitor has a capacitance of \(2 \times 10^{-4}\) farads.
Capacitors are crucial in the given circuit because their ability to store and release charges affects how the voltage across the circuit changes over time.

Ohm's Law

Ohm's Law is a fundamental principle used to understand how electrical circuits behave. It states that the current passing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant.
The simple formula for Ohm's Law is \(I = \frac{V}{R}\), where \(I\) is the current in amperes (A), \(V\) is the voltage in volts (V), and \(R\) is the resistance in ohms (\(\Omega\)).
In the context of the exercise, Ohm's Law helps us determine the current function in the circuit by taking the voltage across the resistor and dividing it by the given resistance (10 \(\Omega\)). This relationship is key to understanding how different elements in the circuit interact with each other and impacts the overall behavior of the system.

RC Circuits

RC circuits are circuits that contain both a resistor (R) and a capacitor (C) connected in series or parallel. These circuits are vital in understanding timing elements in electronics, as they can charge and discharge at predictable rates.
The charging and discharging of a capacitor in an RC circuit follow an exponential pattern. The time it takes to charge or discharge a capacitor is characterized by the time constant \(\tau\), which is calculated as \(\tau = RC\). In the example problem, we found this to be 0.002 seconds, indicating how quickly the capacitor charges to around 63% of the maximum voltage.
RC circuits are also important in filtering applications and creating time delays in electronic devices, making them a pivotal element in various technological applications.