Home
PYQ PracticeFree
JEE Rank Predictor
Answer Key

Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical instrumentHard2 minPYQ_2020

PhysicsHardsingle choice

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

Question diagram: Two resistors 400 Ω and 800 Ω are connected in series across

Options:

Answer:

D

Solution:

Let voltmeter reading is $v$

$\frac{v}{100} \times 400 + (\frac{v}{10000} + \frac{v}{400}) 800 = 6$

$\Rightarrow v + \frac{8 v}{100} + 2 v = 6; \frac{77 v}{25} = 6; v = \frac{150}{77} = 1.95 v$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical instrument

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...