In the given circuit the ammeter A 1 and A 2 are ideal and the ammeter A 3 has a resistance of 1.9 × 10 –3 Ω Ω . If sum of readings of all three meters is given by 
Ampere the value of n will be.
Text Solution
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Sol. As the ammeters A 1 and A 2 are ideal, potential drop across AB and AC are zero. Hence point B and C are at equal potential, so there will be no current through A 3 .
I 3 = 0
Resultant circuit may be drawn as
Applying KVL in the loop ABEFA
– 10 + i 2 5 – 15 i 1 + 20 = 0
⇒ 3i 1 – i 2 = 2 ...
Applying KVL in the loop BCDEB
8 – (i 1 + i 2 ) 3 – i 2 5 + 10 = 0
⇒ 3i 1 + 8i 2 = 18 ...
i 2 = 
Amp, i 1 = 
Amp
Reading of ammeter A 1 , i 1 + i 2 = 
amp
Reading of ammeter A 2 , i 1 = 
amp
i A1 + i A2 + i A3
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