A network of resistore is connected to a 16 V battery with internal resistance of `1Omega,` as shown in (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor. (c ) obtain the voltage drops `V _(AB), V _(BC) and V _(CD)`

Explanation

(i) `R_(AB) = (4)/(2) = 2 Omega`
(ii) `R_(BC) = 1 Omega`
(iii) `R_(CD) = (12 xx 6 )/(12 + 6) = 4 Omega`
Equivalent resistance of given network (between point A and D ) is,
R = `R_(AD) = R_(AB) + R_(BC) + R_(CD)`
= ` 2 + 1 + 4 `
`therefore = 7 Omega`
(b) Current passing through the battery is,
`I = (epsilon)/(R + r) = (16)/(7 + 1) = 2 A `
Between points A and B, two resistances each of `4 Omega` are connected in parallel and so potential difference (p.d.) across them should be equal . thus , `I_(1) xx 4 = I_(2) xx 4 `
`rArr I_(1) = I_(2)`
Now, `I = I_(1) + I_(2) = 2 A rArr I_(1) = I_(2) = 1 A `
Current through 1 `Omega` resistance is I = 2 A
Between point C and D , 12 `Omega and 6 Omega` resistances are connected in parallel and so
`I_(3) xx 12 = I_(4) xx 6`
`rArr I_(4) = 2I_(3)`
Now, I = `I_(3) + I_(4) = I_(3) + 2 I_(3) =3 I_(3) = 2 A `
`" " (because I = 2 A)`
`therefore I_(3) = (2)/(3) A and I_(4) = 2I_(3) = (4)/(3) `A
(C) `V_(AB) = I xx R_(AB) = 2 xx 2 = 4 V`
`V_(BC) = I xx R_(BC) = 2xx 1 = 2 V `
`V_(CD) = I xx R_(CD) = 2xx 4 = 8 V `
Note : Here, terminal voltage of given battery is,
`V = V_(AB) + V_(BC)+ V_(CD ) = 4 + 2 + 8 = 14 `V
But emf of battery is `epsilon = 16 ` V
`rArr` Remaining voltage drop = 16 - 14 = 2 V
is across internal resistance because
Ir =2 `xx 1 = 2 V `
Here, battery is getting discharged and so we get V = `epsilon - Ir = 16 - (2) (I) = 14 ` V