Ohm’s Law Questions and Answers

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Ohm’s law questions and answers are provided here for candidates who are actively preparing for the GATE EC/EE/IN exam. Ohm’s law questions are important for 1 mark. Ohm’s law questions and answers can give a basic understanding of any subject. Any network methodology works based on the application of ohm’s law.

Network reduction techniques will be better understood by using objective-based Ohm’s law MCQ questions and answers. Equivalent resistance, inductance, and capacitance questions come under Ohm’s law questions.

Table of content

1. Ohm’s Law Question 1
2. Ohm’s Law Question 2
3. Ohm’s Law Question 3
4. Ohm’s Law Question 4
5. Ohm’s Law Question 5

Ohm’s Law Question 1

How much resistance is required to limit the current from a 12 V battery to 3.6 mA?

3.3 kΩ
33 kΩ
2.2 kΩ
22 kΩ

Answer: A. 3.3 kΩ

Solution

According to ohm’s law, the current flowing through the element is directly proportional to the potential difference across it.

$I \propto V$

$I = \frac{V}{R}$

Given current I = 3.6 mA

Voltage V = 12 V

Resistance (R) = $\frac{12}{3.6 \times 10^{- 3}} = 3.33 k Ω$

Ohm’s Law Question 2

What is the voltage source for a circuit carrying 2 A of current through a 36 Ω resistor?

1.8 V
18 V
7.2 V
72 V

Answer:  D. 72 V

Solution

Ohm’s law is applicable to passive elements.

Voltage drop (V) across the resistor = Current through element × resistance

Given the current = 2 A

Resistance = 36 Ω

V= I × R = 2 × 36 = 72 V

Only one element is mentioned in the given circuit. Therefore, the voltage drop across the element is equal to the source voltage.

Ohm’s Law Question 3

If you wish to increase the amount of current in a resistor from 120 mA to 160 mA by changing the 24 V source, what should the new voltage setting be?

8 V
320 V
3.2 V
32 V

Answer: D. 32 V

Solution

Ohm’s is applicable for all linear passive elements. According to ohm’s law, the current flowing through the element is directly proportional to the potential difference across it.

$I \propto V$

$\frac{I_{2}}{I_{1}} = \frac{V_{2}}{V_{1}}$

New current $I_{2} = 160 A$

Previous current $I_{1}$ $= 120 A$

Previous voltage $V_{1}$ $= 24 V$

$V_{2} = \frac{I_{2}}{I_{1}} \times V_{1} = \frac{160}{120} \times 24 = 32 V$

Ohm’s Law Question 4

A 120 V lamp-dimming circuit is controlled by a rheostat and protected from the excessive current by a 3 A fuse. To what minimum resistance value can the rheostat be set without blowing the fuse? Assume a lamp resistance of 20 ohms.

40 Ω
4 Ω
2 Ω
20 Ω

Answer: D. 20 Ω

Solution

The bulb is the best example of resistive load. We can apply ohm’s law to the bub circuit.

$T o t a l R e s i s t a n c e = \frac{T o t a l a v a i l a b l e v o l t a g e i n t h e c i r c u i t}{T o t a l c i r c u i t}$

The fuse is blown off when the current exceeds its maximum withstanding capacity. To limit the current in the circuit, additional resistance should keep in the circuit.

Given bulb resistance = 20 Ω

Maximum allowable current = 3A

Available voltage in the circuit is = 120 V

External resistance is R

$R + 20 = \frac{120}{3} = 40$

$R = 40 - 20 = 20 Ω$

Ohm’s Law Question 5

What is the approximate filament resistance of a light bulb if it operates from a 110 V source and 0.6 A of current is flowing?

183 Ω
18.3 Ω
66 Ω
6.6 Ω

Answer:  A. 183 Ω

Solution

The bulb is a resistive load. We can apply ohm’s law.

$F i l a m e n t r e s i s t a n c e = \frac{O p e r a a t i n g v o l t a g e o f t h e b u l b}{C u r r e n t d r a w n b y t h e b u l b} = \frac{110}{0.6} = 183 Ω$

☛ Related Questions:

Ohm’s Law Questions and Answers

Ohm’s Law Question 1

Ohm’s Law Question 2

Ohm’s Law Question 3

Ohm’s Law Question 4

Ohm’s Law Question 5

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