The reading of the ammeter is

0.5 A

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2.25 A

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1.5 A

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0.1 A

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Solution

The correct option is A 2.25 A
The current distributions among the resistors is shown in figure.
Here, $I_{1} = \frac{4.5}{9} = 0.5 A$
As resistor (9,3) and 6 are in parallel so potential across 9 $+$ (potential across 3 $=$ potential across 3)
$\Rightarrow 4.5 + 3 I_{1} = 6 I_{2}$ or $4.5 + 3 (0.5) = 6 I_{2}$
or $I_{2} = \frac{4.5 + 1.5}{6} = 1 A$
So, $I = I_{1} + I_{2} = 0.5 + 1 = 1.5 A$
Equivalent resistance between C and D is $R_{c d} = (1 / 10 + 1 / 12 + 1 / 15)^{- 1} = 4 Ω$
Potential across C and D is $V_{c d} = I R_{c d} = 1.5 \times 4 = 6 V$
Potential across A and D is
$V_{a d} = V_{a b} + V_{b c} + V_{c d} = 6 I_{2} + 2 I + 6 = 6 (1) + 2 (1.5) + 6 = 6 + 3 + 6 = 15 V$
This will be the potential across $20 Ω$ as well as the voltmeter because $20 Ω$ is connected in parallel across total resistance across A and D.
Thus , $I_{4} = \frac{15}{20} = 0.75 A$
Reading in the ammeter $= I + I_{4} = 1.5 + 0.75 = 2.25 A$

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