Question
Download Solution PDFFor the circuit shown below, the voltage across the 11 Ω resistor is given by:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is option3): \(22 \over 3\)V
Concept:
The Kirchoff current law states the algebraic sum of the current at a node is zero
Calculation:
Given
where V is the voltage at node of 1 Ω resistor
By applying KCL
\((12 -V) \over 11\) -V +4 = 0
12 -V -11V = -44
-12V = -56
V = \(56 \over 12\)
= \(14\over 3\)
The Voltage across the 11 Ω resistor is
V = (12 - \(14 \over 3\) )
= \(22 \over 3\) V
Last updated on Jul 1, 2025
-> SSC JE Electrical 2025 Notification is released on June 30 for the post of Junior Engineer Electrical, Civil & Mechanical.
-> There are a total 1340 No of vacancies have been announced. Categtory wise vacancy distribution will be announced later.
-> 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.