Solution:
The current gain \( \beta \) for a common emitter transistor is given by the formula:
\[
\beta = \frac{\Delta I_C}{\Delta I_B}
\]
Where:
\( \Delta I_C = 16 \text{ mA} - 5 \text{ mA} = 11 \text{ mA} \)
\( \Delta I_B = 200 \mu A - 100 \mu A = 100 \mu A \)
Now substitute the values:
\[
\beta = \frac{11 \text{ mA}}{100 \mu A} = \frac{11 \times 10^{-3}}{100 \times 10^{-6}} = 110
\]
Thus, the current gain is \( \boxed{110} \).
The net current flowing in the given circuit is ___ A.
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .