Given ,2Ω and 4Ω resistances are connected in parallel.12 V POTENTIAL difference is applied to those resistances .We need to find the current in the CIRCUIT .When resistances are connected in parallel . Then the resultant resistance is given by ,\bf R_{parallel.Eq}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+.....R parallel.Eq = R 1 1 + R 2 1 + R 3 1 +.....Given that ,\bf R_1=2\;\OMEGA\ ,\;R_2=4\;\OmegaR 1 =2Ω ,R 2 =4Ω$$\begin{lgathered}\implies \bf \dfrac{1}{R_{P.Eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\implies \bf \dfrac{1}{R_{P.Eq}}=\dfrac{1}{2}+\dfrac{1}{4}\\\\\implies \bf R_{P.Eq}=\dfrac{4*2}{4+2}\\\\\implies \bf R_{P.Eq}=\dfrac{8}{6}\\\\\implies \bf R_{P.Eq}=\frac{4}{3}\;\Omega\end{lgathered}$$So we have ,Resistance , R = 4/3 Ω ,Potential Difference , V = 12 VCurrent , I = ? ANow use Ohm's law to find the current .$$\begin{lgathered}\implies \bf V=IR\\\implies \bf 12=I*\dfrac{4}{3}\\\\\implies \bf I=9\;AMPERES\end{lgathered}$$So current in circuit is 9 Amperes