$$\text { Figure } 6 ' 3$$ -If V_S=12 V in Figure 6-3 , what is the value of V_EB if R₃ shorts? A) -3 V B) 8 V C) 6 V D) -8 V

Question

Quizplus · Accepted Answer

When RS1U1B13S1U1B0 shorts, it results in a short circuit between the node U1B1 and U1B0. This means that the voltage at U1B1 will be equal to the voltage at U1B0. Therefore, VS1U1B1EBS1U1B0 is equal to VS1U1B1SS1U1B0 minus the voltage drop across RS1U1B1 (which is equal to the voltage drop across RS1U1B0).

Using voltage division, we can find that the voltage drop across RS1U1B0 is:

VS1U1B1SS1U1B0 * RS1U1B0 / (RS1U1B0 + RS1U1B1) = 12 V * 3 kΩ / (3 kΩ + 2 kΩ) = 7.2 V

Therefore, VS1U1B1EBS1U1B0 = VS1U1B1SS1U1B0 - 7.2 V = 12 V - 7.2 V = 4.8 V

But we have to note that the polarities of the voltages across RS1U1B0 and RS1U1B1 are opposite of each other, because they are both dropping voltage from U1B1 to U1B0. Therefore, the 7.2 V is actually a negative voltage drop across RS1U1B0, which means that the actual polarity of VS1U1B1EBS1U1B0 is negative with respect to U1B0.

Hence, the answer is (D), -8 V.

Figure 6′3\text { Figure } 6 ' 3 Figure 6′3 -If VS=12 V in Figure 6-3

$\text { Figure } 6 ' 3$ -If V_S=12 V in Figure 6-3