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Q.

The equivalent resistance of A and B in the figure below is

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a

\frac{8 r}{7}

b

\frac{7 r}{8}

c

\frac{3 r}{4}

d

r

answer is B.

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Detailed Solution

Concept: A battery is connected to the system of resistors and then applying KCL law to different loops to form equations and then solving it.
It is shown in the figure that the resistance of each resistor is r.
Let’s connect the battery from node A and B as shown in Fig.1,
Question Image

Question Image

The voltage after connecting the battery is V and the total current flowing through the circuit is I. A, B, C, D, E, F, G are the nodes of the circuit. At node A, the current I is split as shown in Figure 1.
Equivalent resistance is given as

R_{eq} = \frac{V}{I}

……………(1)
In the current path, ACEA applying KCL,

I_{1} r = \frac{I - 2 I_{1}}{2} r + (I - 2 I_{1}) r

…………..(2)

3 I = 8 I_{1}

In the path, FACE GF,

V = I_{1} r + \frac{I}{2} r

………….(3)
Putting the value of

I_{1}

from equation (2) in equation (3) .

V = \frac{3}{8} Ir + \frac{I}{2} r = \frac{7}{8} Ir

\frac{V}{I} = \frac{7}{8} r

Putting value of

\frac{V}{I}

in equation (1) ,

R_{eq} = \frac{V}{I} = \frac{7}{8} r

Hence, , the correct Option is 2.

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