Q.In a comman-base connection, a certain transistor has an emitter current of 10mA and collector current of 9.8 mA. Calculate the value of the base current.

For a transistor in the common emitter configuration, the relationship between its three currents ($I_E=I_B+I_C$) leads to two related current-gain ratios: the common-emitter DC current amplification factor $\beta_{dc}=\dfrac{I_C}{I_B}$ (typically 20 to 200 for general-purpose transistors) and the common-base DC current gain $\alpha_{dc}=\dfrac{I_C}{I_E}$ (always a fraction just under 1). Dividing $I_E=I_B+I_C$ through by $I_E$ and rearranging gives the standard relations linking them, $\beta_{dc}=\dfrac{\alpha_{dc}}{1-\alpha_{dc}}$ and equivalently $\alpha_{dc}=\dfrac{\beta_{dc}}{1+\beta_{dc}}$ -- letting either ratio be found directly from the other for a given transistor. …