Three cells of emf 2V,1.8V and 1.5V are connected in series.Their internal resistances are 0.05ohm, 0.7ohm and 1ohm respectively.If the battery is connected to an external resistor of 4ohm via a very low resistance of ammeter, what would be the reading in the ammeter?
Net current I = net emf/net resistance
now net resistance =0.05+4+0.7+1= 5.75 since all resistors are in series
net emf =2+1.8+1.5 =5.3 V
I = 5.3/5.75
0.92 A
hence the current read by ammeter is 0.92 ampere
now net resistance =0.05+4+0.7+1= 5.75 since all resistors are in series
net emf =2+1.8+1.5 =5.3 V
I = 5.3/5.75
0.92 A
hence the current read by ammeter is 0.92 ampere