Question

a cell of emf 3 V and internal resistance 4 ohms is connected to two resistances 10 ohms and 24 ohms joined in parallel. Find the current through each resistance using kirchhoff's laws​

Answer 1

Answer:

Answer is given below....

Answer 2

Answer:

I = 0.2712 A

Explanation:

Given :

  E   = 3 V

  r    = 4 Ω

  R1   = 10 Ω

  R2  = 24 Ω

To Find :  I 1  = Current through R1  = ?,

               I 2 = Current through R2 = ?,

                I = Total current through network

Formula :

i) ΣI = 0 (Kirchhoff’s Junction law)

ii) ΣIR + ΣE = 0 (Kirchhoff’s voltage law)

Solution :

From Kirchhoff’s junction law,

I – I1 – I2 = 0

∴ I = I1 + I2

Apply Kirchhoff’s voltage law for loop in

the circuit containing KLMNOPQK

– I1 R1 – Ir + E = 0

∴ – I1 R1 – (I1 + I2 ) r + E = 0

∴ – 10 I1 – 4 (I1 + I2 ) + 3 = 0

∴ – 10 I1 – 4I1 – 4I2 + 3 = 0

∴ – 14 I1 – 4I2 + 3 = 0

∴ 14 I1 + 4I2 = 3 ... (i)

Apply Kirchhoff’s voltage law for loop

MN′ONM

– I2 R2 + I1 R1 = 0

– 24 I2 + 10 I1 = 0

∴ I1 = 2.4 I2 ... (ii)

From equations (i) and (ii)

14 (2.4 I2 ) + 4I2 = 3

∴ 33.6 I2 + 4I2 = 3

∴ 37.6 I2 = 3

∴ I 2 = $\frac{3}{37.6}$  

I 2 = 0.07979 A ... (iii)

From equations (ii) and (iii)

I 1    = 2.4 × 0.07979

    = 0.1914 A ... (iv)

I    = I1 + I2 = 0.1914 + 0.07979

    = 0.27119

∴ I 1= 0.194 A

I 2   = 0.0798 A

I    = 0.2712 A