Question

1. what is the possible problem if b is equal to 0 v and c is equal to 15 v?

a. transistor is shorted
b. ( r_{b} ) is open
c. ( v_{cc} ) is 0 v
d. transistor is open

Explanation:

Step1: Analyze the circuit operation

In a normal - operating transistor circuit with \(V_{BB} = 15V\) and \(R_{B}=470k\Omega\), if \(R_{B}\) is open, there is no base current \(I_{B}\).

Step2: Determine the transistor state

Since \(I_{B}\approx0\) (because \(R_{B}\) is open), the transistor is in the cutoff state. In the cutoff state, \(I_{C}\approx0\).

Step3: Calculate the voltage at point C

Using Ohm's law \(V = I\times R\), with \(I_{C}\approx0\), the voltage drop across \(R_{C}\) (\(V_{R_{C}}=I_{C}R_{C}\)) is approximately \(0V\). So \(V_{C}=V_{CC}\) (because \(V_{C}=V_{CC}-V_{R_{C}}\)). Given \(V_{CC} = 15V\), \(V_{C}=15V\). And since \(R_{B}\) is open, there is no current through \(R_{B}\), so \(V_{B}\) (measured with respect to ground, assuming ground at the emitter) is \(0V\) (because \(V = IR\) and \(I = 0\) for \(R_{B}\)).

Answer:

B. \(R_{B}\) is open