**Answer: (b) 22.5**

## Explanation

As you can see in the equivalent circuit diagram, it depicts a series combinations of resistances in the circuit. 15Ω, 10Ω, 40Ω, 10Ω, 20Ω, 30Ω..

Let R1 be the equivalent resistance of the series combination of resistance 15Ω, 10Ω.

Similarly R2 be the equivalent resistance of series.

Consider Combination of resistances 40Ω, 10Ω.

Consider Similarly R3 for 20Ω, 30Ω

Then R1, R2, R3 Is given as,

R1=10+15=25Ω

R2=40+10=50Ω

R3=20+30=50Ω

Now, these three resistances R1, R2, R3 are parallel to each other i.e. they are connected in parallel.

Therefore, the equivalent resistance of the parallel combination is given as, 1/Rp = 1/R1 + 1/R2 +1/R3

Where RP = equivalent resistance of the parallel combination

∴ 1/Rp=1/25 + 1/50 + 1/50

= 1/25 + 100/2500

= 1/25 + 1/25

⇒ 1/Rp = 2/25

∴ RP = 25/2 Ω

Now the circuit (simplified circuit) will look like this:

So, the equivalent resistance RP is in series with both resistances 2Ω and 8Ω

∴ Req = Rp+2Ω+8Ω

Where, RP = equivalent resistance (final) of series combination

∴ Req=2 + 25/2 + 8 = 10 + 25/2 = 45/2 =** 22.5Ω**