No links please Two batteries of emfs 5 V and 10 V, their internal resistances are 0.1 Ω and O.5 Ω respectively. They are joined oppositely in series and then connected with a 10 Ω external resistance. What will be the current in the circuit ? Share with your friends Share 0 Decoder_2 answered this Dear stident,As they all are connected in series so,net resistance (R)=0.1+0.5+10=10.7ohmAs battery are connected in series oppositely so,net voltage=5+10=10Vcurrent flowing=net voltagenet resistanceI=1010.7=0.93ARegards 0 View Full Answer
