Question

1. the circuit in figure 10 - 6 has three equal resistors.

(a) if the voltmeter reads + 8.0 v, find the voltage drop across ( r_1 ). ( v_1=)
(b) what is the source voltage? ( v_s=)

1. what basic rules determine if two resistors in a series - parallel combination circuit are connected in series or in parallel?

Explanation:

Step1: Analyze the voltage across \(R_2\) and \(R_3\)

Since \(R_2\) and \(R_3\) are in parallel, the voltage across them is the same. The voltmeter reading \(V = 8.0\space V\), so \(V_{R_2}=V_{R_3}=8.0\space V\).

Step2: Calculate the current through \(R_2\) and \(R_3\)

Let the resistance of each resistor be \(R\). Using Ohm's law \(I=\frac{V}{R}\), for \(R_2\), \(I_{R_2}=\frac{V_{R_2}}{R}=\frac{8}{R}\), for \(R_3\), \(I_{R_3}=\frac{V_{R_3}}{R}=\frac{8}{R}\). The total current through the parallel combination of \(R_2\) and \(R_3\) is \(I = I_{R_2}+I_{R_3}=\frac{8}{R}+\frac{8}{R}=\frac{16}{R}\).

Step3: Calculate the voltage drop across \(R_1\)

The current through \(R_1\) is the same as the total current through the parallel combination of \(R_2\) and \(R_3\) (series - parallel circuit current relationship). Using Ohm's law \(V = IR\), for \(R_1\), \(V_1=I\times R\). Since \(I = \frac{16}{R}\), then \(V_1 = 16\space V\).

Step4: Calculate the source voltage

Using Kirchhoff's voltage law \(V_s=V_1 + V_{R_2}\) (or \(V_{R_3}\) since \(V_{R_2}=V_{R_3}\)). Substituting \(V_1 = 16\space V\) and \(V_{R_2}=8\space V\), we get \(V_s=16 + 8=24\space V\).

Answer:

(a) \(V_1 = 16\space V\)
(b) \(V_s=24\space V\)