# Total resistance of the circuit ?

##### 2 Answers

(a)

(b)

#### Explanation:

(a) Between terminals A and C :

- resistors in arms AB and BC are in series (equivalent
#12 Omega# ) - resistors in arms AD and DC are in series (equivalent
#12 Omega# ) - these two
#12 Omega# resistors are in parallel with the#6Omega# resistance connected directly between A and C

So, the overall equivalent resistance is given by

and thus

(b) Between terminals A and D :

- resistors in arms AB and BC are in series (equivalent
#12 Omega# ) - this is in parallel with the
#6Omega# resistor in AC, giving an equivalent resistance of#(12 times 6)/(12+6)Omega = 4 Omega# - this is in series with the
#6 Omega# resistor in CD, leading to an equivalent of#10 Omega# - Finally, this is in parallel to the
#6Omega# resistor in AD

The overall equivalent resistor

a.

See tips for part b.

#### Explanation:

I would like to offer an alternate method for part a and give some tips for b..

**a.** To get from node A to node C, an electron has 3 possible routes:

#A to C# via the diagonal. Resistance along that path =#6 Omega#

#A to B to C# . Resistance along that path =#12 Omega#

#A to D to C# . Resistance along that path =#12 Omega#

Notice that you can consider paths

So now we have that equivalent resistance,

So the equivalent resistance of the entire network is

**b.** Part b will pretty much require you to use the method posted by Ananda Dasgupta et al.

There is one alternative available if you know the "product over sum" method. Start by choosing the parallel paths

I hope this helps,

Steve