Question

Find the reading of the ideal ammeter connected in the given circuit. Assume that the cells have negligible internal resistance.

• A. 0.8 A
• B. 0.25 A
• C. 1.95 A
• D. 0.05 A

Answer 1

Answer: (C)

In loop ghjig, $-2I_1-6=0\Rightarrow I_1=-3A$
In loop efhge, $-4I_2+2I_1+8=0\Rightarrow I_2^{\prime }=\frac{1}{2}A$
In loop $cdfec,-8I_3+4I_2-4=0\Rightarrow I_3=-\frac{1}{4}A$
In loop $abdca,-10I_4+8I_3+10=0\Rightarrow I_4=\frac{4}{5}A$
Current through ammeter $=I_1+I_2+I_3+I_4$
$=\frac{-60+10-5+16}{20}=\frac{39}{20}A=1.95A$