For the circuit shown in the figure, the current I will be

For the circuit shown in the figure, the current I will be
  1. 0.75 A
  2. 1 A
  3. 1.5 A
  4. 0.5 A

Solution

In the above circuit the cells are connected with opposite polarity it means the net EMF is, E=E1+E2=2+4 V=6 V and, all the resistance is connected in series, R=R1+R2+R3 ⇒I=2+44+1+1=66=1 A.

Asked in: NEET 2020 (Phase 2)

Practice more Current Electricity questions on Aicharya