The current through the 2 Ω resistor is: 

F1 Engineering Arbaz 29-12-23 D6

This question was previously asked in
SSC JE Electrical 9 Oct 2023 Shift 3 Official Paper-I
View all SSC JE EE Papers >
  1. 12A from a to b
  2. 25A from b to a
  3. 1A from a to b 
  4. 10A from b to a

Answer (Detailed Solution Below)

Option 1 : 12A from a to b
Free
RRB JE CBT I Full Test - 23
14.3 K Users
100 Questions 100 Marks 90 Mins

Detailed Solution

Download Solution PDF

Concept

When two branches are connected in parallel, a current is to be found in any one of the branches the current divider rule is used.

F1 Engineering Arbaz 29-12-23 D7

\(I_1=I\times {R_2\over R_1+R_2}\)

\(I_2=I\times {R_1\over R_1+R_2}\)

Calculation:

The current through the 2 Ω resistor is given by:

\(I=36\times {1\over 1+2}\)

I = 12 A from a to b

Additional Information Voltage divider rule:

F1 Engineering Arbaz 29-12-23 D8

The voltage across R2 is given by:

\(V_2=V\times {R_2\over R_1+R_2}\)

The voltage across R1 is given by:

\(V_1=V\times {R_1\over R_1+R_2}\)

Latest SSC JE EE Updates

Last updated on Jun 16, 2025

-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.

-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.

-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31. 

-> Candidates with a degree/diploma in engineering are eligible for this post.

-> The selection process includes Paper I and Paper II online exams, followed by document verification.

-> Prepare for the exam using SSC JE EE Previous Year Papers.

Get Free Access Now
Hot Links: teen patti rummy 51 bonus teen patti chart teen patti wala game teen patti party teen patti master download