Determine the external resistor required to reduce the line voltage from 120 V to 72 V for the operation of a device. The device is rated for 120 V, 100 W.

This question was previously asked in
SSC JE EE Previous Paper 11 (Held on: 24 March 2021 Morning)
View all SSC JE EE Papers >
  1. 48 Ω
  2. 96 Ω
  3. 240 Ω
  4. 144 Ω

Answer (Detailed Solution Below)

Option 2 : 96 Ω
Free
Electrical Machine for All AE/JE EE Exams Mock Test
20 Qs. 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

Concept:

Active power drawn by a device, 

P = V2 / Rd

So, Rd = V2 / P        ...(1)

Where,

R= resistance of the device

V = Voltage rating of the device

Explanation:

Given, device rating is 120 V, 100 W

So, from (1), resistance of device 

Rd = (120)2 / 100 = 144 Ω 

Now let's add external resistor Rex in series with device resistance Rd.

Line voltage of device reduces from 120 to 72 V, so the remaining voltage(120 - 72 = 48 V) should drop across Rex.

Current through circuit, 

Rex = 96 Ω 

Latest SSC JE EE Updates

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.

Hot Links: teen patti lotus teen patti live teen patti download teen patti master list