Current Electricity — Question 22

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

(b) Two resistors of 2 Ω and 3 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{3} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3 + 2}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{5}{6} \\[0.5em] R_p = \dfrac{6}{5} \\[0.5em] \Rightarrow R_p = 1.2 Ω$

Hence, equivalent resistance = 1.2 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 1.2 = 0.6 V

Current through 2 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

0.6 = I x 2 ⇒ I = 0.6 / 2 = 0.3 A

Hence, current through 2 Ω resistor = 0.3 A

Current through 3 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

0.6 = I x 3 ⇒ I = 0.6 / 3 = 0.2 A

Hence, current through 3 Ω resistor = 0.2 A