First slide

Kirchoff's Law

Question

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

Moderate

A

1.95V
B

1.8V
C

2V
D

2.05V

Solution

Let voltmeter measure v volts
$∴ i$ in $10 k Ω$ = $i_{1} = \frac{v}{10000} A$
i in $400 Ω = i_{2} = \frac{v}{400} A$
i in $800 Ω = i = i_{1} + i_{2} = (\frac{v}{10000} + \frac{v}{400}) A$
Applying Kirchhoff’s law for loop ABCDEA

$v + i 800 - 6 = 0 \Rightarrow v + (\frac{v}{10000} + \frac{v}{400}) 800 = 6$
$v = \frac{600}{308}$
$v = 1.95 v o l t s$

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