Question

Fine the equivalent resistance across AB ..

Answer 1

Can you supply a drawing?

Answer 2

There are missing values for 2 out of the 6 resistors in the drawing. Please give these values>

Answer 3

Idiotic me :/ ..

Answer 4

Here it it btw,

Answer 5

The 8 ohm and 2 ohm resistors are in parallel. Let the combined value be R(1).
\[\frac{1}{R(1)}=\frac{1}{2}+\frac{1}{8}\]
The inner 6 ohm resistor and the 5 ohm resistor are also in parallel. Let the combined value be R(2).
\[\frac{1}{R(2)}=\frac{1}{6}+\frac{1}{5}\]
When you have found the values of R(1) and R(2) they should be added, the reason being that these two parallel combinations are in series.
Can you do these calculations and post the result ?

Answer 6

Sure, I would .. But jusy FYI, that another resistor with that 6ohm resistor is 4ohms actually ...

Answer 7

@Aditi_Singh Thanks. So the combined value R(2) is found from
\[\frac{1}{R(2)}=\frac{1}{4}+\frac{1}{6}\]

Answer 8

Hehe.. Obviously , I already changed the values :P ..TY!

Answer 9

Good. So you will soon have a value for the total resistance of the two parallel combinations in series :)

Answer 10

\[\frac{1}{R1} = \frac{5}{8} .. => R1 = \frac{8}{5}\]
\[and.. \frac{1}{R2} = \frac{5}{12} ... => R2 = \frac{12}{5}\]
So, \[R1 + R2 = \frac{8}{5} + \frac{12}{5}\]
\[= \frac{20}{5} = 4 .. \]

Answer 11

Great work! So now to find the resistance between A and B = R(total) use the following expression for 3 ohms, 4 ohms and 6 ohms in parallel:
\[\frac{1}{R(total)}=\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\]

Answer 12

@Aditi_Singh Are you there?

Answer 13

Yup .. So finally.. i get \[R _{e} = \frac{4}{3} ... \] :/

Answer 14

@kropot72 ??

Answer 15

Yes. 1 1/3 ohms is the resistance between A and B. Good work :)