A letter .A. consists of a uniform wire of resisrtance .A. one ohm per cm. The sides of the letter are each 20 cm long and crosspiece in the middle is 10cm long, while the apex angle is `60^@`. Find the resistance of the letter between two ends of the legs A. and E as shown in figure given below.

Explanation

Given resistance of 1 cm of wire = `1 Omega`
`:.` Resistance of 20 cm wire is = `(1 xx 20)Omega`
`20Omega`

As `angleBCD= 60^@` and there is a cross-piece, 10 cm long in the middle
`:. AB = BC = CD = DE = BD = 10 cm` and their resistance
`R_1 = R_2 = R_3 = R_4 = R_5 = 10 Omega`
Now, for resistance of portion BCD

`:.` Their equivalent resistance, `R_s = R_2 + = (10 + 10)Omega = 20Omega`
Now, `R_s and R_5` are in paralle
`:. 1/R_P =1/20 +1/10 =3/20 " or " R_P=20/3Omega`
Now, equivalent resistance between end points A and E = Net resistances of `R_1, R_P and R_4` in series
`:. R= R_1 + R_P + R_4`
`=(10+20/3+10)Omega=80/3Omega`
`R = 26.67 Omega`