The letter A is constructed with a uniform wire of resistance 1 ohm per cm. The sides of the letter are 20 cm long and cross piece in the middle is 10 cm long. The resistance of the letter between the side of the legs is
angle is 60. the triangle formed is an equilateral triangle, so all sides of the triangle are equal
The above 2 sides are in series. So, the resistance directly adds up , that is 10 ohm + 10 ohm= 20 ohms.
This 20 ohm is in parallel with the middle line which is 10 ohm in resistance.
So the triangle is 20/3 ohms
The triangle is in series with the lower lines of A. Add up 10 ohms + 10 ohms + 20/3 ohms. The final answer is 80/3 ohms.
The correct answer is 80/3 ohms. Here is how you find it. See the picture attached below. A triangle is formed in the upper part of the letter A, with all three sides equal to 10cm, therefore the resistance in each of these sides is 10 ohm. The resistance across any two terminals of the triangle will is given as: 2R/3 . Here R=10 So the resistance of the upper part will be 20/3. Now this resistance will appear in series with the lower legs of the letter A, for series we add up all the resisatnces. So Req = 20/3 + 10 + 10 = 80/3 ohm