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

Parallel of 10 k Ω and 400 Ω = 384.61 Ω V 400 Ω = 384.61 384.61 + 800 * 6 = 1.95 V

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

Option 1 - <math> <mn>2</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </math>

Option 2 - <math> <mn>2.05</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </math>

Option 3 - <math> <mn>1.95</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </math>

Option 4 - <math> <mn>1.8</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </math>

2 Views|Posted 5 months ago

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Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 3

Detailed Solution:

Parallel of $10 k Ω$ and $400 Ω = 384.61$ $Ω$

$\begin{array}{r} V_{400 Ω} & = \frac{384.61}{384.61 + 800} * 6 \\ = 1.95 V \end{array}$

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