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Find the equivalent resistance between the terminals F and B in the network shown in the figure below given that the resistance of each resistor is 10 ohm.

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Applying Kirchhoff’s voltage law to loop FGHF, we get,

– 10I1 – 10(I1 – I2) + 10(I – I1) + 10(I – I1) = 0

∴ – 10I1 – 10I1 + 10I2 + 10I – 10I1 + 10I – 10I1 = 0

∴ 201 – 40I1 + 10I2 = 0

∴ 2I – 4I1 + I2 = 0 …………….. (1)

Applying Kirchhoff’s voltage law to loop GABHG, we get,

– 10I2 – 10I2 + 10(I – I2) + 10(I1 – I2) = 0

∴ – 20I2 + 10I – 10I2 + 10I1 – 10I2 = 0

∴ 10I + 10I1 – 40I2 = 0 .

∴ I + I1 – 4I2 = 0 ……………… (2)

Applying Kirchhoff’s voltage law to loop EFHBCDE, we get,

– 10(I – I1) – 10(I – I1) – 10(I – I2) + E = 0

∴ -10I + 10I1 – 10I + 10I1 – 10I + 10I2 + E = 0

∴E = 30I – 20I1 – 10I2 ………….. (3)

From Eq. (1), we get, I2 = 4I1 – 2I …………. (4)

From Eqs. (2) and (4), we get,

I + I1 – 4(4I1 – 2I) = 0

∴ I + I1 – 16I1 + 8I = 0 .

∴ 9I = 15I1 ∴ I1 =

915

I =

35

I …………. (5)

From Eqs. (4) and (5), we get,

I2 = 4(

35

I) – 2I =

125

I – 2I

=

12I−10I5

=

25

I

From Eqs. (3), (5) and (6), we get

E = 30I – 20(

35

I) – 10(

25

I)

= 30I – 12I – 4I = 30I – 16I

∴ E = 14I

If R is the equivalent resistance between E and C,

E = RI

∴ R = 14 Ω

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