Question

For the circuit shown, which one of the following options correctly identifies the Thevenin's equivalent parameters between nodes Y and Z?

The circuit has nodes X, Y, Z, and W. Between X and Y there is a 20 V ideal source (in parallel with a 10 k\(\Omega\) resistor directly connecting X and Y). Between Y and Z there is a 10 k\(\Omega\) resistor. Between Y and W there is a 2 k\(\Omega\) resistor. Between X and W there is an ideal 100 V source (no series resistor in this branch). Between Z and W there is an ideal 20 V source (no series resistor in this branch) and, separately, a 1 k\(\Omega\) resistor.

• A. V_TH = 100 V, R_TH = 10 k\(\Omega\)
• B. V_TH = 140 V, R_TH = 0 \(\Omega\)
• C. V_TH = 100 V, R_TH = 0 \(\Omega\)
• D. V_TH = 140 V, R_TH = 10 k\(\Omega\)

Hint:

Both Y and Z are pinned directly by chains of ideal voltage sources back to the reference node, so their potentials do not depend on any external load; find each node voltage first, then subtract.

Answer 1

The Correct Option is B

Step 1: Take W as the reference (ground) node.
Let \(V_W=0\).
Step 2: Find \(V_X\).
The branch between X and W is a pure ideal \(100\) V source with no series resistance, so it directly fixes
\[ V_X=100\ \text{V} \]
regardless of any current drawn elsewhere in the circuit, since an ideal source's terminal voltage never depends on its current.
Step 3: Find \(V_Y\).
The branch between X and Y is a pure ideal \(20\) V source (in parallel with the \(10\) k\(\Omega\) resistor, which does not affect this voltage relation since the source directly fixes the voltage across both branches). With the source's polarity giving \(V_Y-V_X=20\ \text{V}\):
\[ V_Y=V_X+20=100+20=120\ \text{V} \]
Step 4: Find \(V_Z\).
The branch between Z and W is a pure ideal \(20\) V source (separate from the \(1\) k\(\Omega\) resistor, which is a distinct parallel branch and again does not affect this voltage). With this source's polarity giving \(V_W-V_Z=20\ \text{V}\):
\[ V_Z=V_W-20=0-20=-20\ \text{V} \]
Step 5: Compute \(V_{TH}\).
The Thevenin voltage between Y and Z, taken with the \(+\) reference at Y and \(-\) at Z as marked in the figure, is
\[ V_{TH}=V_Y-V_Z=120-(-20)=140\ \text{V} \]
Step 6: Find \(R_{TH}\).
Both \(V_Y\) and \(V_Z\) are pinned to fixed values entirely by chains of IDEAL voltage sources (with zero series resistance) tracing back to the reference node W. Since neither node's voltage depends even slightly on any current drawn from Y or Z (an ideal source's output voltage is completely independent of its load current), connecting any external load between Y and Z cannot change \(V_Y\) or \(V_Z\) at all. This means the Thevenin resistance seen from Y-Z is
\[ R_{TH}=0\ \Omega \]
Step 7: Rule out the other options.
Options (A) and (D) both include a nonzero \(R_{TH}\) (\(10\) k\(\Omega\)), but this ignores that Y and Z are each rigidly pinned by an all-ideal-source path to ground, making every resistor in the circuit (\(10\) k\(\Omega\) at X-Y, \(10\) k\(\Omega\) at Y-Z, \(2\) k\(\Omega\) at Y-W, \(1\) k\(\Omega\) at Z-W) irrelevant to both \(V_{TH}\) and \(R_{TH}\). Option (C), \(100\) V, corresponds to only \(V_X\) (or \(V_Y\) computed incorrectly without including \(V_Z\)'s contribution), missing the full \(120-(-20)=140\) V difference.
Step 8: Final Answer.
\[ \boxed{V_{TH}=140\ \text{V},\ R_{TH}=0\ \Omega} \]