00:01
In this question we are provided a figure.
00:06
This will be the circuit.
00:07
Let name the resistance.
00:10
This will be r1, r2, r3, r4, r5, r6, r7 and r8.
00:22
And we have to find out the equivalent resistance between a and b.
00:26
From the figure we understand that the resistance r4, r5, r5, r6, sorry, r6, r7 and r8 are in series.
00:48
So the equivalent combination of this series resistance is denoted by r prime will be equal to r4 plus r5 plus r6 plus r7 plus r8.
01:04
That is, substituting the values, we get 2 om plus 2 om plus 4 ome plus 2 ome plus 2 ome so the equivalent resistance of the series r prime is 12 ome then we can redraw the figure now over the circuit reduced like from the figure we understand that resistance r2 and r prime are in parallel the equivalent resistance of the parallel can be found out using let's say r double prime be equal to 1 by r2 plus 1 by r prime...
