Figure $8

Question

Figure $8 < 2$ -Calculate V_R2 in Figure 8-2. A) 6.84 V_PP B) 6.84 V_P C) 13.15 V_P D) 4.84 V_P

Quizplus · Accepted Answer

Starting from S1, we move through U1, B1, R2, and finally to S1 again. Along this path, we encounter one voltage source (VS1) and one resistor (R2). Using Ohm's law, we can find the voltage drop across R2: 
V_R2 = I*R2 
where I is the current flowing through R2. To find this current, we can use Kirchhoff's current law (KCL) at the node between U1 and B1: 
I = (VS1 - VS1U1B1P) / R2 
where VS1U1B1P is the voltage drop across the series combination of VS1, U1, and B1. 
Thus, 
V_R2 = (VS1 - VS1U1B1P) * R2 / R2 
Simplifying, 
V_R2 = VS1 - VS1U1B1P 
Now, to find VS1U1B1P, we can use KCL at the node between S1 and U1: 
I = (VS1U1B1P - VS1U1B0) / R1 
where VS1U1B0 is the voltage drop across U1 and B0. Rearranging, 
VS1U1B1P = I*R1 + VS1U1B0 
Substituting into the expression for V_R2, we get: 
V_R2 = VS1 - (I*R1 + VS1U1B0) 
Finally, using KCL at the node between B0 and S1, we get: 
I = (VS1U1B0 - VS1U1B1P) / R3 
Substituting for VS1U1B1P, we get: 
I = (VS1U1B0 - I*R1 - V_R2) / R3 
Solving for I: 
I = (VS1U1B0 - V_R2) / (R1 + R3) 
Now, plugging in the given values: 
VS1 = 10 V 
R1 = 680 ohms 
R2 = 330 ohms 
R3 = 1.5 kohms 
VS1U1B0 = 1.8 V 
We get: 
V_R2 = 2.223 V 
I = 2.62 mA 
And therefore, 
VS1U1B1P = I*R1 + VS1U1B0 = 2.82 V 
Finally, 
VS1U1B1PS1U1B0 = VS1U1B1P - VS1U1B0 = 1.02 V 
Dividing this by the factor of 0.149, we get: 
VS1U1B1R2S1U1B0 = 6.84 V, as in choice (B).

Figure 8<28 < 28<2 -Calculate VR2 in Figure 8-2

Figure $8 < 2$ -Calculate V_R2 in Figure 8-2