GATE 2020 IN Q.17: solution and answer

GATE 2020 · INQ.17 · 1 mark · NAT

Assuming ideal opamps, the output voltage at $V_{1}$ in the figure shown (in volts) is ____

GATE 2020 IN question 17: Two op-amps. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

$V_{1} = 7 V$

Official answer: 7Source: Official answer key, GATE 2020 IN (IIT Delhi)

Assume ideal op-amps: infinite open-loop gain, infinite input resistance and zero output resistance. Both op-amps have negative feedback. The left one feeds back from $V_{1}$ through a 1 kΩ resistor to its (−) input. The right one feeds back from its output through the 3 kΩ resistor to its (−) input. So each has a virtual short between its inputs, and no current flows into either input terminal.
Virtual short at the left op-amp: its (+) input sits on the 3 V source, so node A, its (−) input, is at $V_{A} = 3 V$ .
Virtual short at the right op-amp: its (+) input sits on the 2 V source, so node B, its (−) input, is at $V_{B} = 2 V$ . The right op-amp holds B at 2 V whatever $V_{1}$ is, so from node A's side, B behaves like a fixed 2 V source.
KCL at node A, with currents in mA and resistances in kΩ. Three currents leave A: through the 1 kΩ to ground, through the 1 kΩ to $V_{1}$ , and through the top 1 kΩ to node B. No current enters the op-amp's input:
$\frac{3 - 0}{1 k Ω} + \frac{3 - V _{1}}{1 k Ω} + \frac{3 - 2}{1 k Ω} = 0$
Solve: $3 + (3 - V_{1}) + 1 = 0 \Rightarrow V_{1} = 7 V$ .
Check at node B, taking currents leaving B (in mA) and calling the right op-amp's output $V_{o 2}$ :
$\frac{2 - 7}{5} + \frac{2 - 3}{1} + \frac{2 - V _{o 2}}{3} = 0 \Rightarrow - 1 - 1 + \frac{2 - V _{o 2}}{3} = 0 \Rightarrow V_{o 2} = - 4 V$
This is finite, so the solution is consistent. The 5 kΩ and 3 kΩ resistors only set the right op-amp's output. They do not affect $V_{1}$ .

KEY IDEA

An ideal op-amp with negative feedback forces its (−) input to the voltage on its (+) input. That fixes node A at 3 V and node B at 2 V. With both ends of every resistor at node A known except $V_{1}$ , KCL at node A alone gives $V_{1}$ . The 5 kΩ and 3 kΩ resistors only decide the right op-amp's output ( $- 4 V$ ).

WATCH OUT FOR

There are three common mistakes. (1) Reading the left stage as a plain non-inverting amplifier with gain $1 + 1/1 = 2$ , which gives $V_{1} = 6 V$ . This ignores the top 1 kΩ, which carries 1 mA from A to B. (2) Treating node B as a virtual ground (0 V) instead of 2 V, which gives $V_{1} = 9 V$ . (3) Writing KCL at the $V_{1}$ node, where the op-amp's output current is unknown, instead of at the op-amp's input node, where the input current is zero.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2020 IN Q.17: solution and answer

Answer and solution

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