# Total resistance of the circuit ?

Apr 14, 2018

(a) $3$ $\Omega$
(b) $3.75$ $\Omega$

#### 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 $6 \Omega$ resistance connected directly between A and C

So, the overall equivalent resistance is given by

$\frac{1}{R} _ e = \left(\frac{1}{12} + \frac{1}{12} + \frac{1}{6}\right) {\Omega}^{-} 1 = \frac{1}{3} {\Omega}^{-} 1$

and thus ${R}_{e} = 3 \Omega$

(b) Between terminals A and D :

• resistors in arms AB and BC are in series (equivalent $12 \Omega$)
• this is in parallel with the $6 \Omega$ resistor in AC, giving an equivalent resistance of $\frac{12 \times 6}{12 + 6} \Omega = 4 \Omega$
• 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 $6 \Omega$ resistor in AD

The overall equivalent resistor

${R}_{e} = \frac{10 \times 6}{10 + 6} \Omega = 3.75 \Omega$

Apr 14, 2018

a. $3 \Omega$.
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 $A \to B \to C$ and $A \to D \to C$ to be parallel. So you can then immediately say (because both resistors have the same resistance: 12=12) that the equivalent resistance of those 2 paths is $\frac{12 \Omega}{2} = 6 \Omega$.

So now we have that equivalent resistance, $6 \Omega$, in parallel with $A \to C$, the diagonal $6 \Omega$.

So the equivalent resistance of the entire network is $\frac{6 \Omega}{2} = 3 \Omega$.

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 $A \to B \to C$ and $A \to C$. The equivalent resistance of that is $\frac{12 \cdot 6}{12 + 6} \Omega$. Then if you add the series $6 \Omega$ between C&D, you have path $A \to B \to C \to D$ and that is parallel with the $6 \Omega$ of path $A \to D$. So you can use product over sum again.

I hope this helps,
Steve