Topic: Analog Circuits
For the circuit given below, the voltage $V_{0}$ across the op-amp output is
(The circuit shows an inverting amplifier with $R_{in}=10~\Omega$ and $R_{f}=50~\Omega$ connecting to node B. Node B is the center of a voltage divider formed by two $10~\Omega$ resistors from the op-amp output $V_{o}$ to ground.)
- A. -9 $V_{i}$
- B. -3 $V_{i}$
- C. -11 $V_{i}$
- D. 9 $V_{i}$
Answer: C. -11 $V_{i}$
Explanation: Due to virtual ground at the inverting terminal (node A), $V_A = 0$.
Applying KCL at node A: $\frac{0-V_A}{50} + \frac{0-V_i}{10} = 0 \Rightarrow V_B = -5V_i$ (Note: Official solution denotes the node as $V_R$ or $V_B$).
Applying KCL at node B: $\frac{V_B}{10} + \frac{V_B-0}{50} + \frac{V_B-V_o}{10} = 0$.
$\Rightarrow \frac{V_B}{5} + 2V_B = V_o \Rightarrow \frac{11V_B}{5} = V_o$.
Substitute $V_B = -5V_i$: $V_o = \frac{11(-5V_i)}{5} = -11V_i$.