Given, voltage gain of first Amplifier, `A_(V_(1))= 10`
Voltage gain of second Amplifier, `A_(V_(2)) = 20`
Input voltage `V_(i) = 0.01 V `
Total voltage gain `A_(v) = ( V_(0))/(V_(i)) = A_(V_(1))xx A_(V_(2))`
`(V_(0))/(0.01) = 10 xx 20 , V_(0)= 2V `.

Correct Answer – B
Such combination is called ” Cascede ” combination
` underset(“net”)(A_(V)) = A_(V1) xx A_(V2) `
`=10 xx 20 = 200`
` :. V_0` = output voltage
` = A_V xx V_i `
` = 200 xx 0.01 `
` = 2 V `

When the amplifiers are Connected in series. The net voltage gain is equal to the product of the gains of the individual amplifiers.
`:. Av = Av^(1) xx Av^(11)`, hear `Av^(1) = 10` , and `Av^(11) = 20`
also `Av = (V_(“output”))/(V_(“input”))`
`:.` we can write
`(V_(“output”))/(V_(“input”)) = Av^(1) xx av^(11) : V_(i n) = 0.01 V`
`V_(out) = V_(i n) xx Av^(1) xx Av^(11) = 0.01 xx 10 xx 20 = 2V`.