a uniform wire of 15 ohm resistance is cut into 3 pieces inthe ratio 1:2:3 and the three pieces are - Science - Electricity

Question

a uniform wire of 15 ohm resistance is cut into 3 pieces inthe
ratio 1:2:3 and the three pieces are connected to form a triangle
acell of 10 v with internal resistance 1 ohm is connected across
the highest of the three resistors calculate the current through
the each part of the circuit

Ruhhi Ralhan · Accepted Answer

Using, $R = ρ \frac{l}{A}$ Given, $l_{1} : l_{2} : l_{3} = 1 : 2 : 3 V = 10 V r= 1 Ω$ Where, $l_{1}, l_{2}, l_{3} are the three lengths of the parts V = voltage applied to the circuit r=internal resistance of the cell$ As, $R α l R_{1} : R_{2} : R_{3} = l_{1} : l_{2} : l_{3} = 1 : 2 : 3 (For same material and of same cross-sectional area) Let R_{1} = x Ω R_{2} = 2 x Ω R_{2} = 3 x Ω R = x + 2 x + 3 x x = 2 5 R_{1} = 2 5 Ω R_{2} = 5 Ω R_{3} = 7 5 Ω$ See the following circuit dig

First find out the main branch current As branch AB and BC are in series and further this series combination is parallel with AC And internal resistance of the cell is in series with the triangular circuit Thus, $R_{e q} = \frac{2 5 \times (5 + 7 5)}{2 5 + (7 5 + 5)} + 1 = 3 08 Ω I = \frac{V}{R_{e q}} = \frac{10}{3 08} = 3 25 A$ Voltage drop across the triangular circuit, $equivalent resistance of triangular circuit: R'= 2 08 Voltage drop, V' = I R' = 3 25 \times 2 08 = 6 76 V$ Now, Current in branch AC, $I_{A C} = \frac{V'}{2 5} = \frac{6 76}{2 5} = 2 704 A$ Current in branch ABC, $I_{A B C} = I - I_{A C} = 0 546 A$

a uniform wire of 15 ohm resistance is cut into 3 pieces inthe ratio 1:2:3 and the three pieces are connected to form a triangle. acell of 10 v with internal resistance 1 ohm is connected across the highest of the three resistors . calculate the current through the each part of the circuit