An ammeter and a voltmeter are connected in series to a battery with an emf $ε = 6.0 V .$ When a certain resistance is connected in parallel with the voltmeter. The reading of the latter decreases $η = 2.0$ times, whereas the reading of the ammeter increases the same number of times. The voltmeter reading after the connection of the resistance is ________ V (Assume ammeter is an ideal)

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1.8 V

5.2 V

3.5 V

2.0 V

answer is A.

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Detailed Solution

Let r = internal resistance of battery G = resistance of voltmeter
Question Image
Initially $V = ε - I r, I = \frac{ε}{r + G}$
$V = \frac{ε G}{r + G} \to (1)$
After the voltmeter is shunted $\frac{V}{η} = ε - \frac{ε r}{r + \frac{R G}{R + G}} (V o l t m e t e r) \to (2)$

and $\frac{ε}{r + \frac{R G}{R + G}} = η . \frac{ε}{r + G} (A m m e t e r) \to (3)$
From (2) & (3) $\frac{V}{η} = ε - \frac{r η ε}{r + G} \to (4)$
From (1) & (4) $\frac{G}{η} = r + G - η r (o r) G = η r$
From (1) $\frac{V}{η} = \frac{ε}{η + 1} = \frac{6}{2 + 1} = 2.0 V$