The reading of the voltmeter (in mV) in the given circuit (assume the voltmeter and the batteries to be ideal) in mV would be _________(upto 2 decimal places).Correct answer is between '181.70,181.90'. Can you explain this answer?

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The reading of the voltmeter (in mV) in the given circuit (assume the voltmeter and the batteries to be ideal) in mV would be _________(upto 2 decimal places).

Correct answer is between '181.70,181.90'. Can you explain this answer?

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The reading of the voltmeter (in mV) in the given circuit (assume the ...

By Kirchhoff's voltage rule.
2V - I₁ x Ω - I x 1Ω- 1V -I x 2Ω =0
⇒1-I₁-3I=0
⇒3I+I₁=1 ......(i)
Again, for the other circuit,
2V-2Ωx(I-I₁)-Ix1Ω-1V-2ΩxI=0
⇒1-5I+2I₁=0
⇒5I-2I₁=1 ......(ii)

Multiplying (i) by 5 and (ii) by 3 and subtracting, we have
11I1 = 2 ⇒ I₁ =2/11A
Therefore, voltage across the 1Ω resistor
=1Ωx2/11A=181.81mV
Since, the voltmeter is parallel to 1Ω resistor, the reading of the voltmeter is 181.81mV.

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The reading of the voltmeter (in mV) in the given circuit (assume the voltmeter and the batteries to be ideal) in mV would be _________(upto 2 decimal places).Correct answer is between '181.70,181.90'. Can you explain this answer?

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