In a CE transistor amplifier, the audio signal voltage across the collector resistance of `2kOmega` is `2V`. If the base resistance is `1kOmega` and the current amplification of the transistor is 100, the input signal voltage is:

Explanation

Here `V_(ce) = 2V, R_(c ) = 2kOmega, beta = 100, R_(1) = 1kOmega i_(beta) = ? V_(B) = ?`
`i_(c )` = Collector currnt`= (V_(ce))/(R_(c )) = (2V)/(2kOmega) = (2V)/(2 xx 10^(-3)Omega) = 1 xx 10^(-3) A`
`i_(B)` = Base current `= (i_(c))/(beta) = (1.0 mA)/(100) = 0.01 mA`
`i_(B) R_(B)` = input signal volatage `= (0.01 mA) (1 kOmega)`
`= (0.1 xx 10^(-3) A) (1 xx 10^(-3) Omega) = 0.01 V = 10 mV`.