In the given circuit, all resistors are 1 k$\Omega$. A 1 mA current source is connected between the top and bottom nodes. A 6 V source connects the midpoints of the two vertical branches as shown. Find the output voltage $V_0$.
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When two identical resistor branches are in parallel, current divides equally.
Always find node voltages first, then apply voltage source constraints.
Step 1: Understand the circuit symmetry.
There are two vertical branches, each having two 1 k$\Omega$ resistors in series.
Hence, resistance of each vertical branch:
\[
R_{branch} = 1k + 1k = 2k\Omega.
\]
These two branches are connected between the same top and bottom nodes. Step 2: Equivalent resistance between top and bottom nodes.
The two 2 k$\Omega$ branches are in parallel:
\[
R_{eq} = \frac{2k \times 2k}{2k + 2k} = 1k\Omega.
\]
A 1 mA current source flows upward.
Hence total voltage between top and bottom nodes is:
\[
V_{ab} = I R_{eq} = (1\,\text{mA})(1k\Omega) = 1\,\text{V}.
\] Step 3: Current division.
Since both branches are identical, current divides equally.
Each branch carries:
\[
I_{branch} = \frac{1}{2} \text{ mA} = 0.5\,\text{mA}.
\]
Voltage drop across each 1 k$\Omega$ resistor:
\[
V = (0.5\,\text{mA})(1k\Omega) = 0.5\,\text{V}.
\] Step 4: Node voltages without the 6 V source.
Let bottom node be reference (0 V).
Then top node is at +1 V.
Midpoint of left branch:
\[
V_L = 0.5\,\text{V}.
\]
Midpoint of right branch (before connecting 6 V source):
\[
V_R = 0.5\,\text{V}.
\] Step 5: Effect of 6 V source.
The 6 V source connects the midpoints.
Given polarity shows left midpoint is 6 V higher than right midpoint:
\[
V_L - V_R = 6.
\]
Since previously both were 0.5 V, the source shifts the right midpoint downward by 6 V relative to left midpoint.
Thus,
\[
V_R = V_L - 6 = 0.5 - 6 = -5.5\,\text{V}.
\] Step 6: Output voltage.
Output $V_0$ is the voltage of right midpoint with respect to ground.
Hence,
\[
V_0 = -5.5\,\text{V}.
\]
