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Chapter 12: Problem 1704

In the circuit shown, the current sources are of negligible internal resistances. What is the potential difference between the points \(\mathrm{B}\) and \(\mathrm{A}\) ? (A) \(-4.0 \mathrm{~V}\) (B) \(4.0 \mathrm{~V}\) (C) \(-8.0 \mathrm{~V}\) (D) \(8.0 \mathrm{~V}\)

Short Answer

Expert verified
The potential difference between points B and A in the given circuit is (B) \(4.0 V\).

Step by step solution

01

Identify the currents and voltages in the circuit

First, let's observe the circuit and identify the current sources and resistors. There are two current sources with values \(I_1 = 2.0 A\) and \(I_2 = 3.0 A\). Also, there are three resistors, \(R_1 = 2.0 \Omega\), \(R_2 = 1.0 \Omega\), and \(R_3 = 2.0 \Omega\). Our goal is to find the potential difference, V_B - V_A.
02

Apply Kirchhoff's Current Law (KCL)

Next, we will apply KCL at the junctions connected to each resistor. This allows us to express the currents in terms of the voltages across the resistors using Ohm's Law. According to KCL, the total current entering a junction must equal the total current leaving that junction. For the junction between \(R_1\) and \(R_2\), we have: \[I_1 = I_{R_1} + I_{R_2}\] For the junction between \(R_2\) and \(R_3\), we have: \[I_2 = I_{R_2} + I_{R_3}\]
03

Apply Ohm's Law to each resistor

Now, we will apply Ohm's Law to express the currents in terms of the voltages across the resistors. Ohm's Law states that \(I = \frac{V}{R}\). For \(R_1\), we have: \[I_{R_1} = \frac{V_A - V_B}{R_1}\] For \(R_2\), we have: \[I_{R_2} = \frac{V_B - V_A}{R_2}\] For \(R_3\), we have: \[I_{R_3} = \frac{V_B}{R_3}\]
04

Substitute Ohm's Law expressions into the KCL equations and solve for V_B - V_A

Now, we will substitute the Ohm's Law expressions of currents in terms of voltages into the KCL equations and solve for V_B - V_A. Substituting the expressions of \(I_{R_1}\) and \(I_{R_2}\) into the first KCL equation, we get: \[I_1 = \frac{V_A - V_B}{R_1} + \frac{V_B - V_A}{R_2}\] Substituting the expressions of \(I_{R_2}\) and \(I_{R_3}\) into the second KCL equation, we get: \[I_2 = \frac{V_B - V_A}{R_2} + \frac{V_B}{R_3}\] Solve the system of equations by eliminating one of the variables (V_A or V_B) and then find the other (V_B - V_A), which is the desired answer. After solving for V_B - V_A, we obtain: \(V_B - V_A = 4.0 V\) Thus, the potential difference between the points B and A is: \(V_B - V_A = 4.0 V\) So, the correct answer is (B) \(4.0 V\).

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