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SarkariResultOutBMS · 2024

Previous Year Question

Obtain the voltage v in the branch shown in Fig. Q6 for i2 = −2 A.
Fig. Q6 dependent voltage source branch with vₓ = 15i₂ and a 10 V source
Fig. Q6 from the original paper, cleanly cropped with complete polarity and source labels.
  1. 20 V
  2. −20 V
  3. 30 V
  4. −10 V

Correct answer: B

Explanation

The controlling current is signed: i2 = −2 A. Hence vx = 15i2 = −30 V. The two source voltages add algebraically in the marked polarity, so v = vx + 10 = −20 V.

Formula

vx = 15i2
v = vx + 10
v = 15(−2) + 10 = −20 V

Step-by-step solution

Series Source Polarity and KVL explanatory diagram

Correct Answer

B — −20 V

The dependent source evaluates to −30 V and the series 10 V source gives a net branch voltage of −20 V.

What This Question Is Really Testing

This is a sign-convention question involving a current-controlled dependent voltage source and KVL. The important point is to keep the negative sign of the controlling current and then combine source voltages according to the polarity marks shown in Fig. Q6.

Concept Foundation

A diamond-shaped source is a dependent source. Its value is determined by another circuit variable. Here the source voltage is vx = 15i2. The +/− marks define the reference polarity. If the calculated source voltage is negative, the actual polarity is opposite to that reference; the negative sign must not be discarded.

Step 1 — Evaluate the Dependent Source

Use the controlling current exactly as given, including its sign.

i2 = −2 A
vx = 15i2
vx = 15(−2) = −30 V

Step 2 — Relate the Two Series Source Voltages

The branch voltage v is measured from the upper + reference to the lower − reference. Both source reference polarities are shown + at the top and − at the bottom, so their signed voltages add.

v = vx + 10 V

Step 3 — Substitute and Simplify

v = −30 + 10
v = −20 V

Therefore the correct option is B.

Why This Method Works

KVL uses algebraic source voltages, not just magnitudes. A negative value of vx automatically accounts for the actual polarity being opposite to the marked reference.

Independent Verification / Cross-Check

Bottom → top rise = 10 + (−30) = −20 V

The same signed sum reproduces the requested branch voltage, confirming the result.

Concept Extension

  • A dependent source remains active during normal circuit analysis.
  • Its value must first be written in terms of the controlling current or voltage.
  • Negative controlled-source values are physically valid and indicate reversed actual polarity.

Exam Strategy

For dependent-source MCQs: evaluate the controlling expression first → keep its sign → then apply KVL/KCL. Never replace a signed current by its magnitude.

Other options

A uses the wrong sign for the controlled source. C uses only the magnitude 30 V and ignores the 10 V series source. D does not correctly combine the two signed source voltages.

Common mistake

Replacing i₂ = −2 A by |i₂| = 2 A. That reverses the dependent-source result and leads to a wrong branch voltage.

Exam tip

Write the dependent-source value beside the source before starting KVL; it prevents sign errors.

Quick method

15(−2) = −30 V; then add +10 V → −20 V.