In the network shown, resistance between terminals a and b is X. Then

X - 8 = 4

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X - 8 = 0

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X + 8 = 16

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X + 8 = 20

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Solution

The correct options are
A $X - 8 = 0$
C $X + 8 = 16$
Loop equation I (via 2 $Ω$ )
$2 i_{1} + 4 (i_{1} - i_{2}) + 6 (i_{1} - i_{2} - i_{3}) = V$ or $12 i_{1} - 10 i_{2} - 6 i_{3} = V$
$∴ 6 i_{1} - 5 i_{2} - 3 i_{3} = \frac{V}{2}$ ........(1)
Loop equation II (via 4 $Ω$ )
$4 (I - i_{1}) + 8 (I - i_{1} + i_{2}) + 12 (I - i_{1} + i_{2} + i_{3}) = V$
$6 I - 6 i_{1} + 5 i_{2} + 3 i_{3} = \frac{V}{4}$ ...............(2)
(1)+(2): $6 I = \frac{3 V}{4}$ or $\frac{V}{I} = \frac{24}{3} = 8 Ω$

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The equation of the bisectors of the angle between lines represented by equation $4 x^{2} - 16 x y - 7 y^{2} = 0$ is

Factorise: $1 - 16 (x - 2 y)^{2}$ .

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