Question

Considering the ideal op-amp circuit in the figure 2: (i) Find the value of the resistance R such that the current Is is zero which means the resistance seen from Vs is infinite, $R_s = \infty$. (ii) If the output saturation voltage for both op-amps are \pm12V, determine the range of Vs in the linear range. 

          Considering the ideal op-amp circuit in the figure 2:
(i) Find the value of the resistance R such that the current Is is zero which means the resistance seen from Vs is infinite, $R_s = \infty$.
(ii) If the output saturation voltage for both op-amps are \pm12V, determine the range of Vs in the linear range.
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Considering the ideal op-amp circuit in the figure 2: (i) Find the value of the resistance R such that the current Is is zero which means the resistance seen from Vs is infinite, Rs = ∞. (ii) If the output saturation voltage for both op-amps are ±12V, determine the range of Vs in the linear range.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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please answer Considering the ideal op-amp circuit in the figure 2: iFind the value of the resistance R such that the current Is is zero which means the resistance seen from Vs is infinite,Rs=o. ii If the output saturation voltage for both op-amps are 12V,determine the range of Vs in the linear range R MW 40 k WW 20 k WW 10 kQ MM 10kQ RS=T Vo Vs Figure2
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Transcript

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00:02 Hello student, here due to virtual ground potential at the inverting terminal is zero.
00:07 So, applying kirchhoff's law in this branch we have zero minus ix into 20 into 10 to the power 3 minus minus of 3 equals to zero.
00:20 So, ix equals to 1 .5 into 10 to the power minus 4 ampere.
00:27 So, ix equals to 1 .5 into 10 to the power minus 4 ampere.
00:40 Now, let's assume the current in the circuit in this circuit is i1 and current in the circuit is i2.
00:51 So, i1 equals to equals to 4 divided by 10 into 10 to the power 3 ampere.
01:05 So, which is equals to 4 into 10 to the power minus 4 ampere and i2 equals to 6 divided by 30 into 10 to the power 3 ampere.
01:20 So, which is equals to 2 into 10 to the power minus 4 ampere...
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