Question

In the circuit shown below, the value of the output voltage \(V_0\) is \(\_\_\_\_\_\) V.

The circuit has two 1 \(\Omega\) resistors in series along the top, forming three nodes: a left node, a middle node, and the output node (\(V_0\) is measured between the output node and the bottom rail). Two 2 \(\Omega\) resistors connect the left node to the bottom rail, one 2 \(\Omega\) resistor in series with a 4 V source connects the middle node to the bottom rail, and another 2 \(\Omega\) resistor in series with a 4 V source connects the output node to the bottom rail. The output terminals are open, so no external load draws current.

• A. 2
• B. 3
• C. 4
• D. 8

Hint:

Combine the two parallel 2 \(\Omega\) resistors at the left node into 1 \(\Omega\) first, note that the node between each 2 \(\Omega\) resistor and its 4 V source is fixed at 4 V, then write the node equations (or use superposition) to find the output node's voltage.

Answer 1

The Correct Option is B

Step 1: Label the nodes and simplify the parallel resistors.
Call the left node \(N_1\) (where the two plain 2 \(\Omega\) resistors and the first 1 \(\Omega\) resistor meet), the middle node \(N_2\) (where the 2 \(\Omega\) resistor with the first 4 V source connects), and the output node \(N_3\) (where \(V_0\) is measured, and where the 2 \(\Omega\) resistor with the second 4 V source connects). Take the bottom rail as the reference, 0 V. Since \(N_1\) has two 2 \(\Omega\) resistors going to ground side by side, combine them first:
\[ 2\ \Omega \parallel 2\ \Omega = \frac{2\times2}{2+2} = 1\ \Omega \]
So \(N_1\) connects to ground through a single 1 \(\Omega\) resistor.

Step 2: Use the fixed voltage below each source branch.
Each 4 V source sits below its own 2 \(\Omega\) resistor, with its + terminal facing up toward the resistor and its - terminal at the ground rail. Since these are ideal sources, the node between each 2 \(\Omega\) resistor and its source is always 4 V above ground. So the current leaving \(N_2\) through its branch is \(\dfrac{V_2-4}{2}\), and the current leaving \(N_3\) through its branch is \(\dfrac{V_3-4}{2}\). Because the \(V_0\) terminals are open, all the current reaching \(N_3\) must flow down through this branch.

Step 3: Write the node equations (KCL, current leaving each node sums to zero).
At \(N_1\):
\[ \frac{V_1}{1} + \frac{V_1-V_2}{1} = 0 \ \Rightarrow\ 2V_1=V_2 \]
At \(N_2\):
\[ \frac{V_2-V_1}{1} + \frac{V_2-V_3}{1} + \frac{V_2-4}{2} = 0 \]
At \(N_3\):
\[ \frac{V_3-V_2}{1} + \frac{V_3-4}{2} = 0 \]

Step 4: Solve the equations.
From \(N_1\), \(V_1=V_2/2\). Substitute into the \(N_2\) equation and multiply through by 2:
\[ 2(V_2-V_1)+2(V_2-V_3)+(V_2-4)=0 \ \Rightarrow\ 5V_2-2V_1-2V_3=4 \]
Using \(V_1=V_2/2\):
\[ 5V_2-V_2-2V_3=4 \ \Rightarrow\ 4V_2-2V_3=4 \ \Rightarrow\ V_3=2V_2-2 \]
From the \(N_3\) equation, multiply through by 2:
\[ 2(V_3-V_2)+(V_3-4)=0 \ \Rightarrow\ 3V_3=2V_2+4 \ \Rightarrow\ V_3=\frac{2V_2+4}{3} \]
Set the two expressions for \(V_3\) equal:
\[ 2V_2-2 = \frac{2V_2+4}{3} \ \Rightarrow\ 6V_2-6=2V_2+4 \ \Rightarrow\ 4V_2=10 \ \Rightarrow\ V_2=2.5\ \text{V} \]
Then
\[ V_3 = 2(2.5)-2 = 3\ \text{V} \]

Step 5: Check the other options.
Option (A), 2 V, equals the contribution of just the source nearest the output acting alone, if you forget to add in the effect of the other source. Options (C) and (D) are too large because they do not account for how much the resistor ladder divides down each source's effect before it reaches the output.

Final Answer:
\[ \boxed{V_0 = 3\ \text{V}} \]