Step 1: Find the collector current (\( I_C \)). The collector current (\( I_C \)) is determined by: \[ I_C = \frac{V_{CC}}{R_C}, \] where: - \( V_{CC} = 1 \, \text{V} \), - \( R_C = 1 \, \text{k}\Omega = 1000 \, \Omega \). Substitute the values: \[ I_C = \frac{1}{1000} = 1 \, \text{mA}. \]
Step 2: Relate base current (\( I_B \)) and collector current (\( I_C \)). The current gain (\( \beta \)) of the transistor is given by: \[ \beta = \frac{I_C}{I_B}. \] Rearrange to solve for \( I_B \): \[ I_B = \frac{I_C}{\beta}. \] Substitute \( I_C = 1 \, \text{mA} = 10^{-3} \, \text{A} \) and \( \beta = 100 \): \[ I_B = \frac{10^{-3}}{100} = 10 \, \mu \text{A}. \]
Final Answer: The base current is: \[ \boxed{I_B = 10 \, \mu \text{A}}. \]
If $ \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p $, then $ 96 \log_e p $ is equal to _______