Answer:
[tex]i_A[/tex] = 3 A
[tex]i_B[/tex] = 1.8 A
[tex]i_C[/tex] = 1.2 A
Explanation:
Combined resistance of the resisters B and C,
[tex]\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}[/tex] [Since [tex]R_1[/tex] and [tex]R_2[/tex] are attached in parallel]
[tex]\frac{1}{R'}=\frac{1}{4}+\frac{1}{6}[/tex]
[tex]\frac{1}{R'}=\frac{10}{24}[/tex]
R' = 2.4 Ω
Since, R' and resistor A are in series therefore, total resistance of both the resistors = 2 + 2.4
= 4.4 Ω
Now the current flowing in the circuit 'i' = [tex]\frac{E}{R+r}[/tex]
Here, E = E.M.F. of the cell
R = Total resistance of all resistors attached
r = Resistance of the cell
'i' = [tex]\frac{9}{2.4+0.6}[/tex]
i = 3 A
Current flowing in resistor A = 2 A
And Current in combined resistance of B and C = 3 A
Current flow in the resistors B and C will be in the ratio of C : B,
[tex]\frac{i_C}{i_B}=\frac{B}{C}[/tex]
[tex]\frac{i_C}{i_B}=\frac{4}{6}[/tex]
[tex]\frac{i_C}{i_B}=\frac{2}{3}[/tex]
Therefore, [tex]i_C[/tex] = [tex]\frac{2}{(2+3)}\times 3[/tex] = 1.2 A
[tex]i_B=\frac{3}{(2+3)}\times 3[/tex] = 1.8 A
