Question

1. determine the base current ( i_b ).

given: ( \beta = 200, v_{be} = 0.7 v )
a. ( 30 mu a )
b. ( 23 mu a )
c. ( 30 ma )
d. ( 23 ma )

Explanation:

Step1: Apply KVL to base - emitter loop

Using Kirchhoff's Voltage Law (\(KVL\)) in the base - emitter loop: \(V_{BB}=I_{B}R_{B}+V_{BE}\).
We can re - arrange the formula for \(I_{B}\) as \(I_{B}=\frac{V_{BB}-V_{BE}}{R_{B}}\).

Step2: Substitute the given values

Given \(V_{BB} = 3V\), \(V_{BE}=0.7V\), and \(R_{B}=100\times10^{3}\Omega\).
Substitute these values into the formula: \(I_{B}=\frac{3 - 0.7}{100\times10^{3}}\).
First, calculate the numerator: \(3 - 0.7=2.3\).
Then, \(I_{B}=\frac{2.3}{100\times10^{3}}A\).
Since \(1A = 10^{6}\mu A\), \(I_{B}=\frac{2.3}{100\times10^{3}}\times10^{6}\mu A\).
\(I_{B}=23\mu A\).

Answer:

B. \(23\ \mu A\)