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

Let voltmeter reading is v 100 v × 400 + ( 10000 v + 400 v ) 800 = 6 ⇒ v + 100 8 v + 2 v = 6 ; 25 77 v = 6 ; v = 77 150 = 1.95 v

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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical instrumentHard2 minQB

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

Options:

Answer:

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

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