Question

In the given op-amp circuit, the non-inverting terminal is grounded. The input voltage is 2 V applied through 1 k$\Omega$. The feedback resistor is 1 k$\Omega$. The output is connected to a 2 k$\Omega$ load to ground and also through a 2 k$\Omega$ resistor to the op-amp output. Find the output voltage $V_0$ and currents $I_1$, $I_0$, and $I_x$. 
 

Hint:

In an inverting amplifier, first find $V_0$ using gain formula. Then apply KCL at the output node to compute additional load currents.

Answer 1

Correct Answer: 3

Step 1: Understanding the Question
This question asks for the analysis of an inverting operational amplifier circuit with a specific load configuration. We need to determine the output voltage $V_0$ and three different currents, $I_1$, $I_0$, and $I_x$, as marked in the circuit diagram.
Step 2: Key Formula or Approach
The primary concepts for solving this are the virtual ground principle for an inverting op-amp and Kirchhoff's Current Law (KCL) at the various nodes.
Virtual Ground: Since the non-inverting terminal ($V_+$) is grounded (0 V), the inverting terminal ($V_-$) will also be at 0 V.
KCL: The sum of currents entering a node must equal the sum of currents leaving the node.
Step 3: Detailed Explanation
Finding the input current and $I_1$:
Due to the virtual ground, the voltage at the inverting terminal ($V_-$) is 0 V. The input voltage of 2 V is applied through a 1 k$\Omega$ resistor to this node. The input current flowing towards the node is:
\[ I_{in} = \frac{V_{in} - V_-}{R_{in}} = \frac{2\,\text{V} - 0\,\text{V}}{1\,\text{k}\Omega} = 2\,\text{mA} \] Since an ideal op-amp has zero input current, this entire input current must flow through the feedback path. The current $I_1$ is defined as the current in the feedback resistor, so:
\[ I_1 = I_{in} = 2\,\text{mA} \] Finding the output voltage $V_0$:
The current $I_1$ flows from the inverting node (0 V) to the output node $V_0$ through the 1 k$\Omega$ feedback resistor. Using Ohm's Law for the feedback resistor:
\[ V_0 - V_- = -I_1 \times R_f \] \[ V_0 - 0 = -(2\,\text{mA}) \times (1\,\text{k}\Omega) = -2\,\text{V} \] So, the output voltage is $V_0 = -2\,\text{V}$.
Finding the load current $I_0$:
The current $I_0$ flows through the 2 k$\Omega$ load resistor connected between the output node $V_0$ and ground.
\[ I_0 = \frac{V_0}{R_{load}} = \frac{-2\,\text{V}}{2\,\text{k}\Omega} = -1\,\text{mA} \] The negative sign indicates the current flows upwards from ground to the output node. The magnitude of the current is $1\,\text{mA}$.
Finding the op-amp output current $I_x$:
The current $I_x$ is the total current supplied by the op-amp's internal output stage. This current reaches the output node $V_0$ and then splits into the feedback current ($I_1$) and the load current ($I_0$). Applying KCL at the output node $V_0$ (sum of currents leaving = 0):
\[ -I_x + I_1 + I_0 (\text{magnitude}) = 0 \] \[ I_x = I_1 + I_0 (\text{magnitude}) = 2\,\text{mA} + 1\,\text{mA} = 3\,\text{mA} \] Step 4: Final Answer
The calculated values are:
\[ V_0 = -2\,\text{V} \] \[ I_1 = 2\,\text{mA} \] \[ I_0 = 1\,\text{mA} \quad (\text{magnitude}) \] \[ I_x = 3\,\text{mA} \]