Question

the equivalent resistance between point p and q

Answer 1

Hope this will help :)

Answer:- 1.5 R

Answer 2

Answer:$\frac{3}{2}R$

Explanation:

The equivalent resistance in when resistors are connected in series is:

$Rs=R_1+R_2+R_3+...$

The equivalent resistance in when resistors are connected in parallel is:

$Rp=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...$

Refer to the figure attached below

R4 and R5 are in parallel connection:

$R_{45}=[\frac{1}{R4}+\frac{1}{R5}]^{-1}$

$R_{45}=[\frac{1}{2R}+\frac{1}{2R}]^{-1}=[\frac{2}{2R}]^{-1}=R$

So, we can replace combination of R4 and R5 by single resistor $R_{45} =R$

Now, R2, R3 and  $R_{45}$ are in series. We can replace these with equivalent resistor:

$Rs=R2+R3+R_{45}=R+R+R =3R$

Now Rs and R1 are in connected in parallel between points P and Q, so equivalent resistance is $R_{eq}=[\frac{1}{R1}+\frac{1}{Rs}]^{-1}=[\frac{1}{3R}+\frac{1}{3R}]^{-1}=\frac{3}{2}R$

Hence, the equivalent resistance between points P and Q is $\frac{3}{2}R$