Two resistors 400   Ω and 800   Ω are connected in series across a 6 ∨ battery.…

Two resistors 400 Ω  and  800 Ω are connected in series across a 6∨ battery. The potential difference measured by a voltmeter of 10 kΩ across 400 Ω resistor is close to:
  1. 2 V
  2. 1.8 V
  3. 2.05 V
  4. 1.95 V

Solution

Let voltmeter reading is v

v100×400+v10000+v400 800=6

⇒   v+8v100+2v=6 ; 77v25=6 ; v=15077=1.95 v

Asked in: JEE Main 2020 (03 Sep Shift 2)

Practice more Current Electricity questions on Aicharya