JEE Main20203 Sep 2020Evening ShiftPhysicsCurrent ElectricityActual
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:
Options
- A2   V
- B1 . 8   V
- C2 . 05   V
- D1 . 95   V
Correct answer
D. 1 . 95   V
Step-by-step solution
Let voltmeter reading is v v 100 × 400 + v 10000 + v 400   800 = 6 ⇒       v + 8 v 100 + 2 v = 6   ;   77 v 25 = 6   ;   v = 150 77 = 1 . 95   v