If three resistors of resistance 2 ,Omega,4 , | vedclass

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(A) For resistors connected in parallel,the equivalent resistance $R_{eq}$ is given by the formula:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} +

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$\frac{20}{19} \,\Omega$

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(A) For resistors connected in parallel,the equivalent resistance $R_{eq}$ is given by the formula:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$Given $R_1 = 2 \,\Omega$,$R_2 = 4 \,\Omega$,and $R_3 = 5 \,\Omega$.Substituting the values:$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5}$Taking the least common multiple $(LCM)$ of $2, 4,$ and $5$,which is $20$:$\frac{1}{R_{eq}} = \frac{10 + 5 + 4}{20} = \frac{19}{20} \,\Omega^{-1}$Therefore,the total resistance is:$R_{eq} = \frac{20}{19} \,\Omega$

If three resistors of resistance $2 \,\Omega$,$4 \,\Omega$ and $5 \,\Omega$ are connected in parallel,then the total resistance of the combination will be:

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