Question

What is the combined resistance of each of the networks between A and B shown in fig. ?

Answer 1

(a) Between A andB,resistances 5Ω and 3 Ω are connected in series

·. R_s =5+3=8Ω
This series combination of resistances 5Ω and 3Ω is connected in parallel with the resistance 80
:. total resistance between A and Bis given as:

R = `[1/8 + 1/8]^-1` = 4Ω

(b) Bet ween A and B ,paralle combination of resistances 9Ω and 18Ω is connected in series with resistance 2Ω.
Parallel resistance of 9Ω and 18Ω is :

R_p = `[1/9 + 1/18]^-1` = 6Ω

:. total resistance between A and B is:
R=6+2=8Ω

(c) The situation consists of three two ohm resistors connected in series between CEFD and their combination in parallel with the fourth 2 ohm resistor between C and D.

Therefore, series combination gives, 2 + 2 + 2 = 6 Ω.
This 6Ω resistor is connected in para lie I to the fourth 2Ω n resistor, therefore equivalent resistance between C and D,

`1/"R" = 1/6 + 1/2 = 2/3 Ω`

.·.R=l.5Ω

Now, between A and B the resistance Rand two 2Ω resistors are connected in series.
Therefore, equivalent resistance between A and B is

R_total = 2+2 +1.5 =5.5 Ω