The resistors of resistances 2 Omega,4 Omeg | vedclass

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Question

(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}

Vedclass · Accepted Answer

$\frac{8}{7} \ \Omega$

Vedclass · Answer

(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 $R_3 = 8 \ \Omega$.Substituting the values: $\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8}$.Finding a common denominator,which is $8$: $\frac{1}{R_{eq}} = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{4+2+1}{8} = \frac{7}{8} \ \Omega^{-1}$.Therefore,$R_{eq} = \frac{8}{7} \ \Omega$.

The resistors of resistances $2 \ \Omega$,$4 \ \Omega$,and $8 \ \Omega$ are connected in parallel. The equivalent resistance of the combination will be:

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