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Q.
Find the reading of the ideal ammeter connected in the given circuit. Assume that the cells have negligible internal resistance.
Current Electricity
Solution:
In loop ghjig, $-2 I_{1}-6=0 \Rightarrow I_{1}=-3 A$
In loop efhge, $-4 I_{2}+2 I_{1}+8=0 \Rightarrow I_{2}^{\prime}=\frac{1}{2} A$
In loop $c d f e c,-8 I_{3}+4 I_{2}-4=0 \Rightarrow I_{3}=-\frac{1}{4} A$
In loop $a b d c a,-10 I_{4}+8 I_{3}+10=0 \Rightarrow I_{4}=\frac{4}{5} A$
Current through ammeter $=I_{1}+I_{2}+I_{3}+I_{4}$
$=\frac{-60+10-5+16}{20}=\frac{39}{20} A =1.95\, A$
