Verified Solution • 2025–26 Session moderate

QJEEM20Phc828: JEE Main 2020 Physics - Class 12 JEEM20Phc828 Solution: Two resistors 400 and 800 are connected in series across a 6 V…

Two resistors 400 Ω\Omega and 800 Ω\Omega are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 k Ω\Omega across 400 Ω\Omega resistor is close to :

Multiple Choice Options

  • A 2.05 V
  • B 1.95 V
  • C 2 V
  • D 1.8 V

Step-by-Step Verified Solution

JEE Main 2020 (Online) 3rd September Evening Slot Physics - Current Electricity Question 216 English Explanation


i=6800+400×10000400+10000i = {6 \over {800 + {{400 \times 10000} \over {400 + 10000}}}}

i=6800+40000104i = {6 \over {800 + {{40000} \over {104}}}}

i=6800+384.61=61184.61=0.00506i = {6 \over {800 + 384.61}} = {6 \over {1184.61}} = 0.00506

V400 = i1 ×\times 400

= (104400+104)i×400\left( {{{{{10}^4}} \over {400 + {{10}^4}}}} \right)i \times 400

= (104400+104)(0.00506)×400\left( {{{{{10}^4}} \over {400 + {{10}^4}}}} \right)\left( {0.00506} \right) \times 400

= 1.95 V

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring