JEE Main202430 Jan 2024Evening ShiftPhysicsCurrent ElectricityActual
Two resistance of 100 Ω and 200 Ω are connected in series with a battery of 4 V and negligible internal resistance. A voltmeter is used to measure voltage across 100 Ω resistance, which gives reading as 1 V . The resistance of voltmeter must be _______ Ω .
Correct answer
0
Step-by-step solution
Voltage across 200 Ω = 4 - 1 = 3 V . Therefore, current through the battery, 3 200 Now, equivalent resistance of R V & 100 Ω , = R v 100 R v + 100 For voltmeter, R v 100 R v + 100 × 3 200 = 1 ⇒ 3 R v = 2 R v + 200 ⇒ R v = 200 Ω