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Nepal Eng. Council15th Exam · 2083 BS

Jan 4, 2027 · 100 Marks · Pass: 50%

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Chapter 8: Software Engineering & OOAD
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Q-abf24553

ACtE08 · Chapter 8Demo2 · Set 1🏛️ Past Licensure QuestionID: abf24553

The equivalent resistance of a $1\,\Omega$ resistor connected in parallel with a $3\,\Omega$ resistor is:

Multiple Choice Options

A
0.50 Ω
B
0.75 Ω
✓ Correct Option
C
0.33 Ω
D
4.00 Ω

Detailed Technical Solution & Explanation

For two resistors $R_1$ and $R_2$ connected in parallel, the equivalent resistance $R_{eq}$ is calculated using the formula:

R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2}

Given $R_1 = 1\,\Omega$ and $R_2 = 3\,\Omega$ :

R_{eq} = \frac{1 \times 3}{1 + 3} = \frac{3}{4} = 0.75\,\Omega

Alternatively, using conductance ( $G = 1/R$ ):

G_{eq} = G_1 + G_2 = \frac{1}{1} + \frac{1}{3} = 1 + 0.333 = 1.333\text{ S}

R_{eq} = \frac{1}{G_{eq}} = \frac{1}{1.333\text{ S}} = 0.75\,\Omega

Option Analysis:

•

Option A (0.50 Ω): Incorrect; represents two

1\,\Omega

resistors in parallel (

1/2 = 0.5\,\Omega

).

•

Option B (0.75 Ω): Correct;

\frac{1 \times 3}{1 + 3} = \frac{3}{4} = 0.75\,\Omega

.

•

Option C (0.33 Ω): Incorrect; represents

1/3\,\Omega

.

•

Option D (4.00 Ω): Incorrect; represents the series combination (

1 + 3 = 4\,\Omega

).

Reference: Demo2 · Set 1NEC Licensure Syllabus

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