If a battery of emf 8V and netgligible internal resistance in connected between terminals P and Q of the circuit shown in fig 5.225 calculate the current through 2.5 Omega resistance and hence calculate the equivalent resistance of the circuit.

Explanation

Let the current drawn by the circuit from battery be `I = (I_1 + I_2)` and currenty distribution be as shwon in fig

Applying kirchhoff's voltage law in mesh
BCMKB,
`10(I_1 - I_3) - 15(_3 +I_4) - 25I_3 = 0` (i)
Applying Kirchhoff's voltage law in mesh
ABKLGHA,
`25I_3 - 25I_4 - 5I_2= 0 (ii)`
Applying Kirchhoff's Voltage law in mesh
`LMCDEFGL`,
`15(I_3 + I_4) - 3(I_2 -I_4) + 25I_4 = 0 (iii)`
Applying Kirchhoff's voltage law in mesh
PJHFNQXYP,
`5I_2 +3 (I_2 - I_4)- 8= 0 (iv)`
From above equations,
`I-1= I-2 = IA`
`I_3 = 0.2A`
`I_4 = 0`
Total current drawn from battery is ` I =I_1 + I_2 = 2A`. Therefore,
equivalent resistance is
`R = (E )/(I) = (8)/(2) = 4Omega`