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.

Question

Suresh said, "I did it yesterday."
Which one of the following options is the correct form of this sentence in indirect speech?

Answer 1

Since this circuit has two independent voltage sources and is built entirely from resistors, we can solve it faster with superposition: work out $V_0$ due to each 4 V source acting alone, with the other source turned off (an ideal voltage source set to zero becomes a plain wire), then add the two results.

Label the nodes as before: $N_1$ is the left node (with the two parallel 2 $\Omega$ resistors to ground, which combine to $1\ \Omega$), $N_2$ is the middle node (with the first 4 V source's branch), and $N_3$ is the output node (with the second 4 V source's branch, where $V_0$ is measured).

Case 1: only the source in the $N_3$ branch is active (the $N_2$ branch source is shorted, so that branch becomes a plain 2 $\Omega$ resistor to ground).

From $N_2$, there are now two 2 $\Omega$ paths to ground: one direct, and one through the 1 $\Omega$ resistor into $N_1$ and then $N_1$'s 1 $\Omega$ to ground (giving $1+1=2\ \Omega$). These two 2 $\Omega$ paths are in parallel:

$$ 2\ \Omega \parallel 2\ \Omega = 1\ \Omega $$

From $N_3$, this $1\ \Omega$ equivalent at $N_2$ is reached through the 1 $\Omega$ resistor between $N_3$ and $N_2$, giving $1+1=2\ \Omega$ as the resistance the source sees outside its own 2 $\Omega$ arm. The full loop resistance for the 4 V source is then $2+2=4\ \Omega$, so the current it drives is:

$$ I = \frac{4\text{ V}}{4\ \Omega} = 1\text{ A} $$

This current flows into the $2\ \Omega$ equivalent resistance at $N_3$, giving that source's contribution to $V_0$:

$$ V_0^{(1)} = 1\text{ A}\times2\ \Omega = 2\text{ V} $$

Case 2: only the source in the $N_2$ branch is active (the $N_3$ branch source is shorted, so that branch becomes a plain 2 $\Omega$ resistor to ground).

From $N_2$, the source now sees two parallel paths: one straight into $N_1$ (1 $\Omega$) and then to ground (1 $\Omega$), totaling $2\ \Omega$; the other into $N_3$ (1 $\Omega$) and then to ground through the now-plain 2 $\Omega$ resistor, totaling $3\ \Omega$. These combine as:

$$ 2\ \Omega \parallel 3\ \Omega = \frac{2\times3}{2+3} = 1.2\ \Omega $$

The full loop resistance for this source is its own $2\ \Omega$ plus this $1.2\ \Omega$, giving $3.2\ \Omega$, so the current it drives is:

$$ I = \frac{4\text{ V}}{3.2\ \Omega} = 1.25\text{ A} $$

This current splits between the two parallel paths at $N_2$ in inverse proportion to their resistance (current divider rule). The share going toward $N_3$, through the $3\ \Omega$ path, is:

$$ I_{N_3} = 1.25\text{ A}\times\frac{2\ \Omega}{2\ \Omega+3\ \Omega} = 1.25\times0.4 = 0.5\text{ A} $$

This current then passes through the $2\ \Omega$ resistor at $N_3$ to ground, giving that source's contribution to $V_0$:

$$ V_0^{(2)} = 0.5\text{ A}\times2\ \Omega = 1\text{ V} $$

Combine both contributions:

$$ V_0 = V_0^{(1)}+V_0^{(2)} = 2\text{ V}+1\text{ V} = 3\text{ V} $$

Let's summarize:

Superposition lets us handle each source with a plain series-parallel resistor reduction, instead of solving three simultaneous node equations at once.
The source nearest the output contributes 2 V by itself, and the other source, after being divided down by the ladder, contributes 1 V.
Adding them gives $V_0=3$ V, matching option (B).

$$ \boxed{V_0 = 3\text{ V}} $$

Answer 2