Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $$R_p = 1 \Omega$$ as shown in the figure. An external resistance of $$R_e = 2 \Omega$$ is connected via the sliding contact. The electric current in the circuit is :

step 1: split the potentiometer

Total Rp=1 Ω, slider at middle → each half:

top half = 0.5 Ω
bottom half = 0.5 Ω

step 2: understand connections

Let left end = A, right end = B, slider = C

Between A and C:
• direct path = 0.5 Ω
• another path = 2 Ω (external resistor)

So between A and C → parallel:

$$R_{AC}=(0.5\times2)/(0.5+2)$$

RAC=1/2.5=0.4 Ω

step 3: from C to B

Only one resistor:

RCB=0.5 Ω

step 4: total resistance

Series combination:

Req=0.4+0.5=0.9 Ω

step 5: current

Battery = 0.9 V

I=V/R=0.9/0.9=1 A