Question: b) Find $i_1$ and $i_2$ in the summing op amp circuit in Figure 5. (6 marks)
2 V
5 k\Omega
2.5 k\Omega
1 V
a
b
10 k\Omega
i_0
i_1
i_2
2 k\Omega
v_0
+
1
Figure 5: Summing OP amp.
top-left: 2 V source on the left connected to the summing node through a 5 k\Omega resistor; 1 V source below-left connected to the same node through a 2.5 k\Omega resistor.
top-right: feedback path from the op-amp output to the inverting input through a 10 k\Omega resistor; labels a, i_0, i_1, and i_2 are near the op-amp/output node.
center: op-amp triangle with inputs labeled a (upper, inverting) and b (lower, non-inverting), with b tied to ground; output on the right.
bottom-right: 2 k\Omega resistor from the output node to ground with v_0 marked positive at the top and negative at the bottom.
1. **Problem Statement:**
Find the currents $i_1$ and $i_2$ in the summing operational amplifier circuit shown in Figure 5.
2. **Circuit Description and Assumptions:**
- The op-amp is ideal, so the voltage at the inverting input node (node $a$) is virtually at ground potential (0 V) because the non-inverting input $b$ is grounded.
- The currents $i_1$ and $i_2$ flow through the resistors connected to the voltage sources $2\text{ V}$ and $1\text{ V}$ respectively into the summing node.
3. **Known Values:**
- Voltage sources: $V_1 = 2\text{ V}$, $V_2 = 1\text{ V}$
- Resistors: $R_1 = 5\text{ k}\Omega$, $R_2 = 2.5\text{ k}\Omega$
- The summing node voltage $V_a = 0\text{ V}$ (virtual ground)
4. **Formulas:**
Current through a resistor is given by Ohm's law:
$$i = \frac{V_{source} - V_{node}}{R}$$
Since $V_a = 0$, currents are:
$$i_1 = \frac{2 - 0}{5\times10^3} = \frac{2}{5000}$$
$$i_2 = \frac{1 - 0}{2.5\times10^3} = \frac{1}{2500}$$
5. **Calculations:**
$$i_1 = \frac{2}{5000} = 0.0004\text{ A} = 0.4\text{ mA}$$
$$i_2 = \frac{1}{2500} = 0.0004\text{ A} = 0.4\text{ mA}$$
6. **Interpretation:**
Both currents $i_1$ and $i_2$ are $0.4\text{ mA}$ flowing into the summing node.
**Final answers:**
$$i_1 = 0.4\text{ mA}, \quad i_2 = 0.4\text{ mA}$$