Equating XXX currents at XXXXX and XXXXXX XXXX,
To achieve XXXXXXX we will be XXXXXXX,
Then XXX result will XX,
XXX let's XXXX XXXX the XXXXXXXXXXXX amplifier XXXXXXX.
XXX circuit XXXXX XXXX,
Same current rule applies XXX XXX XXXXXXX XXXXX XXXXX Voltage at point X (VA) will be XXXXX XX XXXXXXX at point X(VB).
Now XXXX on! what's XX should XX?
XX XX w the XXXXXXX drop XXXXXX XX resistor, XXXXX is,
Let's XXX XXXX XXXXXXX analysis,
The current XX is what XX XXXX XXXXXXX at XXX output side XXXXXXX of XXXX input impedance of the XX-XXX,
XXXXXXXX the current XX XXXXX and output side,
Solving XXXXXXXXXXXXX,
So, output voltage XXXXXXX,
Now XX achieve XXX XXXXXXXXXXXX XXXXXXXXX we should XX the XXXXXXXXX,
Now XXX XXXXXX XXXXXXX,
Differential operation achieved!!
Let's XXXX onto XXX problem,
XXXXXXX XXX circuit, it's just a XXXXXXX XXXXXXXXX and we just XXX the XXXXXXXX value XXXXX in XXX problem into the XXXXXXXX XX XXX,
where,
XXXXXXX these values,
XXXXXX XXXX XX -40V.
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