Three resistances 2 Omega, 3 Omega and 4 Omeg | vedclass

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(B) According to Ohm's law,$V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.Since the resistors are connect

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$6: 4: 3$

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(B) According to Ohm's law,$V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.Since the resistors are connected in parallel,the potential difference $V$ across each resistor is the same.Therefore,$I = V/R$,which implies $I \propto 1/R$.Let the currents through the $2 \Omega, 3 \Omega$,and $4 \Omega$ resistors be $I_1, I_2$,and $I_3$ respectively.Then,$I_1 : I_2 : I_3 = \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$.To simplify this ratio,multiply each term by the least common multiple $(LCM)$ of $2, 3$,and $4$,which is $12$.$I_1 : I_2 : I_3 = (\frac{1}{2} 	imes 12) : (\frac{1}{3} 	imes 12) : (\frac{1}{4} 	imes 12) = 6 : 4 : 3$.

Three resistances $2 \Omega, 3 \Omega$ and $4 \Omega$ are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be

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