A wire of resistance 18 Ω is divided into three equal parts. These parts are connected in side of triangle, the equivalent resistance of any two corners of triangle will be

Question

A wire of resistance 18 Ω is divided into three equal parts. These parts are connected in side of triangle, the equivalent resistance of any two corners of triangle will be (A) 18 Ω  (B) 9 Ω  (C) 6 Ω  (D) 4 Ω

Tardigrade Admin · Accepted Answer

To find equivalent resistance across BC, AB and AC are in series R'=6+6=12 Ω 12 Ω and 6 Ω is in parallel Total resistance, R=(12× 6/12+6) R=(72/18)=4Ω

Q. A wire of resistance $18 Ω$ is divided into three equal parts. These parts are connected in side of triangle, the equivalent resistance of any two corners of triangle will be

18 $Ω$

9 $Ω$

6 $Ω$

4 $Ω$

To find equivalent resistance across BC, AB and AC are in series
$R^{'} = 6 + 6 = 12 Ω$
$12 Ω$ and $6 Ω$
is in parallel Total resistance,
$R = \frac{12 \times 6}{12 + 6}$
$R = \frac{72}{18} = 4Ω$

Q. A wire of resistance 18Ω is divided into three equal parts. These parts are connected in side of triangle, the equivalent resistance of any two corners of triangle will be

18 Ω

9 Ω

6 Ω

4 Ω