The total current supplied to the circuit as shown in figure by the 5 V battery is A. [Figure: a circular wire loop with a 5 V battery on its left side, splitting it into an upper arc (node P) and a lower arc (node Q). Inside, a star centre O connects to two points on the upper arc through 5 Ω and…

Question

The total current supplied to the circuit as shown in figure by the 5 V battery is A. [Figure: a circular wire loop with a 5 V battery on its left side, splitting it into an upper arc (node P) and a lower arc (node Q). Inside, a star centre O connects to two points on the upper arc through 5 Ω and 5 Ω, and to the lower arc through 2.5 Ω. A further 5 Ω resistor on the right joins the upper arc to the lower arc.]

Accepted Answer

2. Both upper 5 Ω star arms end on the same arc P, so they are in parallel: 2.5 Ω. Star path P→ O→ Q: 2.5+2.5=5 Ω. This is in parallel with the right 5 Ω: R_eq=2.5 Ω. I=5/(2.5)=2 A.

Step-by-step solution