Question

In the arrangement of resistances as shown, what is the equivalent resistance between $A$ and $B$ ?

• A. $4r$
• B. $3r$
• C. $2.5r$
• D. $r$

Answer 1

Answer: (D)

We can simplify the circuit diagram

And we can see that between A and C terminal 2r and 2r resistances are parallel combination and we can calculate by using the formmula $(R_{eq}=\frac{R_1\times R_2}{R_1+R_2})$
Hence $\frac{2r\times 2r}{2r+2r}=r$
Similarly we can find the parallel combination between B and C terminal $\frac{2r\times 2r}{2r+2r}=r$
Again we can see that between C and D terminal r and r resistances are parallel combination and again we calculate by using formula $(R_{eq}=\frac{R_1\times R_2}{R_1+R_2})$
hence we get $\frac{r\times r}{r+r}=\frac{r}{2}$

And now we get a simplifying circuit

Hence we can understand that it follows the wheatstone bridge concept and balance the following equation which is $\frac{P}{Q}=\frac{R}{S}$
where P = Q = R = S = r Ω hence no current will be drawn from $\frac{r}{2}$ Ω resistance i.e
Equilent resistance is $\frac{2r\times 2r}{2r+2r}=r$
(Because r and r are in series combination so $r+r=2r$ )