Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.

(a) Draw a labelled diagram of the arrangement.

(b) Calculate the current in each resistor.

(a) Labelled diagram of the arrangement is shown below:

(b) Two resistors of 4 Ω and 6 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R$p} = \dfrac{1}{4} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{3 + 2}{12} \\[0.5em] \dfrac{1}{Rp} = \dfrac{5}{12} \\[0.5em] Rp = \dfrac{12}{5} \\[0.5em] \Rightarrow R_p = 2.4 Ω

Hence, equivalent resistance = 2.4 Ω

Current I = 0.5 A

Potential Difference V = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

V = 0.5 x 2.4 = 1.2 V

Current through 4 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

1.2 = I x 4
⇒ I = 1.2 / 4 = 0.3 A

Hence, current through 4 Ω resistor = 0.3 A

Current through 6 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

1.2 = I x 6
⇒ I = 1.2 / 6 = 0.2 A

Hence, current through 6 Ω resistor = 0.2 A