The terminals of a 18 ~V battery with an internal resistance of 1.5 are connected to a circular wire of resistance 24 at two points distant at one quarter of the circumference of a circular wire. The current through the bigger arc of the circle will be

Step-by-step solution

Since, the terminals of battery is connected at one quarter distance. Hence, ratio of lengths between two parts will be aligned l₁ l₂ &= 1 3 l₁: l₂ &=1: 3 Since, R l R₁: R₂ &=1: 3 R₁ &= 24 1+3 =6 R₂ &=24 3 1+3 =18 aligned Since, potential difference across R₁ and R₂ is same, so they are connected in parallel. Equivalent resistance of R₁ and R₂ , aligned R^ =& R₁ R₂ R₁+R₂ =& 6 18 6+18 =45 Total resistance, R_ T &=R^ +r &=4.5+1.5 [ r=1.5 ] &=6 aligned Current, I= V R_ T = 18 6 =3 ~A According to current division rule