Final answer:
The voltage gain of the first stage is 50, the input resistance of the first stage is 10000 Ω, and the current gain from the input of the first stage to the collectors of the second stage is 10000.
Step-by-step explanation:
To find the voltage gain of the first stage of a BJT differential amplifier, we will use the transconductance (gm) and the load resistance (Rc). The transconductance gm is given as Ic/Vt, but since we are given re, we have gm = 1/re because re = Vt/Ie and Ic ≈ Ie for a BJT. Therefore, gm is 1/100 Ω = 0.01 S. The voltage gain Av is given by gm multiplied by the load resistance, so Av = gm * Rc = 0.01 S * 5000 Ω = 50.
The input resistance of the first stage, Rin, is essentially the base resistance looking into the transistor, which can be approximated by β * re since β is the current gain of the transistor and re is the intrinsic emitter resistance. With β = 100 and re = 100 Ω, Rin = β * re = 10000 Ω.
For the current gain from the input of the first stage to the collectors of the second stage, since both stages are BJT differential amplifiers with a current gain approximately equal to the β of the transistors, the overall current gain would be the product of the β of the first and second stage transistors, which is β2 = 1002 = 10000.