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. 

Ohm’s Law Question 1 

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

  1. 3.3 kΩ 
  2. 33 kΩ 
  3. 2.2 kΩ 
  4. 22 kΩ 

Answer: A. 3.3 kΩ 


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





Given current I = 3.6 mA 

Voltage V = 12 V 

Resistance (R) = 123.6×103=3.33 k

>12/(3.6×10-3) = 3.33 kΩ

Ohm’s Law Question 2

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

  1. 1.8 V
  2. 18 V
  3. 7.2 V
  4. 72 V

Answer: D. 72 V


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  

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?

  1. 8 V
  2. 320 V
  3. 3.2 V
  4. 32 V

Answer: D. 32 V


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. 



I2I1= V2V1

>I2/I= V2/V1

New current I2=160 A

>I2=160 A

Previous current I1


=120 A

>=120 A=120 A

Previous voltage V1


=24 V

>=24 V=24 V

V2=I2I1×V1=  160120×24=32 V

>V2=I2/I1×V1=  160/120×24=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. 

  1. 40  
  2. 4  
  3. 2  
  4. 20  

Answer: D. 20


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

Total Resistance= Total available voltage in the circuitTotal circuit

>Total Resistance = Total available voltage in the circuit/Total circuit

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=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? 

  1. 183  
  2. 18.3  
  3. 66  
  4. 6.6  

Answer: A. 183 Ω 


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

Filament resistance= Operaating voltage of the bulbCurrent drawn by the bulb= 1100.6=183 

>Filament resistance= Operating voltage of the bulb/Current drawn by the bulb= 110/0.6=183Ω 

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