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

In loop  efhge,  −4I2 +2I1+8 =0  ⇒In  loop cdfec,  −8I3 +4I2−4 =0  ⇒ In  loop abdca,  −10I4 +8I3+10 =0 in loop abjia -10I4-6+8-4+10=0 I4=810=45A, I3=-14A, I2=12A, I1=-3ACurrent through ammeter =I1+I2+I3+I4=−60+10−5+1620  =3920 A=1.95  A