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Question

In figure the voltmeter and ammeter shown are ideal. Then voltmeter and ammeter readings, respectively, are:
155740_8adbaea195574857af5a729e7f107dd3.png

A
32.978 V, 3 A
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B
3 V, 0.236 A
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C
120 V, 4.198 A
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D
120 V, 3 A
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Solution

The correct option is B 3 V, 0.236 A

Ideal voltmeter means it has resistance, current will not flow in the branch of a voltmeter.
Ideal ammeter means it has zero resistance.So that it can read maximum current.
Given VAVC=6V (1)
VBVC=1.50V (2)
Subtract equation (1) and (2)
VAVB=4.5V
So reading of voltmeter =VAVB=4.5V
Now we can assume that EMF6V has an internal resistance of 10Ω and 1.5V has 5Ω.
Both are in parallel combination.
Eeq=E1r2=E2r1r1+r2
=6×5+1.5×1010+5
=30+1515
=3V
req=r1r2r1+r2
=10×515
=5015=103Ω
Now the circuit will becomes

Equal current will flow through the same resistance.
Using Kirchhoff's law in loop (1)
i2×63+103×i=0
6i2+10i3=3
(3+103)i=3
193i=3
[i=919A]
Ammeter will read the reading=i2
=919×2
=938A=0.236A
Hence B is correct option.

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