In the figure ammeter (I) reads a current of 10mA, while the voltmeter reads a potential difference of 3V. What does ammeter (II) (in mA) read? The ammeter are identical, the internal resistance of the battery is negligible. (Consider all ammeters and voltmeters as non ideal)

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answer is 6.66.

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Detailed Solution

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Given $\frac{1 - 0}{r_{A}} = 10 m A$
$r_{A} = \frac{1}{10 \times 10^{- 3}} = 100 Ω$
Nodal $\frac{x - 1}{r_{A}} + \frac{x - 4}{100} + \frac{x - 0}{100} = 0$
$\frac{x - 1}{100} + \frac{x - 4}{100} + \frac{x - 0}{100} = 0$
$x = \frac{5}{3}$
Current in Ammeter (II) $= \frac{\frac{5}{3} - 1}{100} A = 6.66 m A$

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