Question

In Figure \(10.39, R_{L}=Z_{0}\) and \(R_{g}=Z_{0} / 3\). The switch is closed at \(t=0\). Determine and plot as functions of time \((a)\) the voltage across \(R_{L} ;(b)\) the voltage across \(R_{g} ;(c)\) the current through the battery.

Short answer

Answer: The expressions for the voltage across \(R_{L}\), the voltage across \(R_{g}\), and the current through the battery are \(V_{R_L} = \frac{3}{4}V_{battery}\), \(V_{R_g} = \frac{1}{4}V_{battery}\), and \(I = \frac{3V_{battery}}{4Z_{0}}\), respectively.

Step-by-step solution

Identify constants and circuit configuration

First, let's identify the given values and circuit configuration. We have \(R_{L}=Z_{0}\), \(R_{g}=Z_{0}/3\), and the switch is closed at \(t=0\). The circuit is a simple series circuit consisting of three resistors and a battery.

Use Kirchhoff's Laws and Ohm's Law to find the current and voltages

We will start by using Kirchhoff's Voltage Law (KVL) and Ohm's Law to find the expressions for the current passing through the circuit and the unknown voltages across the resistors. Kirchhoff's Voltage Law states that the sum of the voltages around a closed loop is equal to zero. In our case, we have: \(V_{battery} - V_{R_L} - V_{R_g} = 0\) Ohm's Law states that the voltage across a resistor is equal to the current passing through it times its resistance: \(V_{R_L} = I \cdot R_{L}\) \(V_{R_g} = I \cdot R_{g}\)

Solve for the unknowns

Now, we can substitute the expressions for \(V_{R_L}\) and \(V_{R_g}\) from equation (2) and (3) into the general KVL equation (1): \(V_{battery} - I \cdot R_{L} - I \cdot R_{g} = 0\) Rearranging and solving for \(I\): \(I = \frac{V_{battery}}{R_{L}+R_{g}}\) Substituting given values for \(R_{L}\) and \(R_{g}\) in terms of \(Z_{0}\): \(I = \frac{V_{battery}}{Z_{0}+\frac{Z_{0}}{3}} = \frac{3V_{battery}}{4Z_{0}}\) Now that we have the expression for the current, we can substitute it back into the equations for the voltages across the resistors: \(V_{R_L} = I \cdot R_{L} = \frac{3V_{battery}}{4Z_{0}} \cdot Z_{0} = \frac{3}{4}V_{battery}\) \(V_{R_g} = I \cdot R_{g} = \frac{3V_{battery}}{4Z_{0}} \cdot \frac{Z_{0}}{3} = \frac{1}{4}V_{battery}\)

Plot the functions of time

Now that we have the expressions for the voltages across the resistors and the current passing through the battery, we can plot these functions as a function of time. To plot the voltage across \(R_L\) as a function of time: \(V_{R_L}(t)= \frac{3}{4}V_{battery}u(t)\), where \(u(t)\) is the unit step function, representing the switch being closed at \(t=0\). To plot the voltage across \(R_g\) as a function of time: \(V_{R_g}(t)= \frac{1}{4}V_{battery}u(t)\) To plot the current through the battery as a function of time: \(I(t) = \frac{3V_{battery}}{4Z_{0}}u(t)\) To conclude, we have found the expressions for the voltage across \(R_{L}\), the voltage across \(R_{g}\), and the current through the battery as functions of time.

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering that connects voltage (V), current (I), and resistance (R) in a circuit. It is expressed as \(V = I \cdot R\). This means the voltage across a resistor is directly proportional to the current flowing through it and its resistance. In the given exercise, we used Ohm's Law to determine the voltages across resistors \(R_L\) and \(R_g\) in a series circuit. By knowing these relationships, we can solve for unknown variables, such as current or voltage across a component. Remember, Ohm's Law only applies to linear, ohmic materials, where the resistance remains constant over varying voltages and currents.

Voltage across resistors

In a circuit, the voltage drop across each resistor indicates how much voltage is "used up" by that resistor as current flows through it. The total battery voltage is divided among all the resistors in the circuit. In our exercise, we have two resistors, \(R_L\) and \(R_g\), in a series circuit.

• For\(R_L\), the voltage drop is \(V_{R_L}= I \cdot R_{L} = \frac{3}{4}V_{battery}\).
• For \(R_g\), it is \(V_{R_g}= I \cdot R_{g} = \frac{1}{4}V_{battery}\).

These expressions demonstrate how the voltage splits in series circuits, in proportion to their resistance. Understanding voltage drop is crucial for circuit analysis, especially in ensuring components function within their voltage ratings.

Series circuit analysis

Series circuits are characterized by a single path through which current flows, passing through each resistor in sequence. One key property of series circuits is that the current remains constant through each component. This feature is vital when performing circuit analysis, allowing us to find the unknown values, such as voltage and resistance. In our case, applying Kirchhoff’s Voltage Law, we derived: \[V_{battery} - V_{R_L} - V_{R_g} = 0\]Combining it with Ohm's Law, we concluded that the total voltage is split across the resistors, proportional to their resistance. In series circuits, ensuring all components share the current helps in quantifying total circuit impedance and optimizing circuit functionality.

Transient response in circuits

Transient response refers to how a circuit responds to changes over time, especially after switching events. When the switch is closed at \(t=0\), the circuit reaches a new steady state, displaying transient behavior initially. For our series circuit, the transient response analysis can be visualized using step functions, like:

• The voltage across \(R_L\) as \(V_{R_L}(t)= \frac{3}{4}V_{battery}u(t)\).
• The voltage across \(R_g\) as \(V_{R_g}(t)= \frac{1}{4}V_{battery}u(t)\).
• The current is given by \(I(t) = \frac{3V_{battery}}{4Z_{0}}u(t)\).

The step function \(u(t)\) indicates these values change instantaneously as the switch is closed, exemplifying the concept of transient response.