Question

Find the Equivalent resistance between A and B.

Answer 1

Answer:

So we get the equivalent by the formula, 1Req=1R1+1R2+1R3+.... Now in the top wire in between the points A and B we have 2 resistances which are in series. So the equivalent resistance between the points A and B will be 11R15.

Explanation:

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Answer 2

Answer:

Equivalent resistance of the network is $\frac{5}{3}\: \Omega$

Explanation :

Given,

Circuit diagram of resistor connect in a network like cube.

To find:

Equivalent resistance of the network

Solution:

Formula that will be used:

1) When n resistance are in series:

$\green{\boxed{R_{eq}=R_1+R_2+R_3+...+R_n}}$

2) When n resistance are in parallel

$\green{\boxed{\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}}}$

Traditional method to solve this type of problem is to convert this network in a simplified network.

See in attachment.

1) All the resistance connected with point A (marked a,b and c) are in parallel.

2) All the resistance connected with point B(marked d,e and f) are also in parallel.

3) All the in between resistances (marked as g,h,i,j,k and l) are in parallel connection.

All 3 cases are connected in series.

Now,

Step 1:

Apply the formula of parallel connection for case 1

$\frac{1}{r_x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \\ \\ \frac{1}{r_x} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} \\ \\ r_x = \frac{1}{3} \:\Omega$

Step 2:

Apply the formula of parallel connection for case 2

$\frac{1}{r_y} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f} \\ \\ \frac{1}{r_y} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\ \\ r_y = 1 \:\Omega$

Step 3:

Apply the formula of parallel connection for case 3

$\frac{1}{r_z} = \frac{1}{g} + \frac{1}{h} + \frac{1}{i} + \frac{1}{j} + \frac{1}{k} + \frac{1}{l} \\ \\ \frac{1}{r_z} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} +\frac{1}{2}\\ \\ r_z = \frac{1}{3} \:\Omega$

Step 4:

As discussed above all three arrangements of parallel resistors are in series.

Thus

$R_{AB} = r_x + r_y + r_z \\ \\ R_{AB} = \frac{1}{3} +1+ \frac{1}{3} \\ \\ R_{AB} = \frac{1+3+1}{3} \\ \\ R_{AB} = \frac{5}{3}\: \Omega$

So,

Equivalent resistance of the network is $\frac{5}{3}\: \Omega$