Two resistors of resistances 2 Omega and 6 Om | vedclass

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(A) The current $I$ flowing through the battery is given by the formula $I = \frac{E}{R_{eq} + r}$,where $E$ is the emf,$R_{eq}$ is the equivalent ext

Vedclass · Accepted Answer

$1 	ext{ A}$

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(A) The current $I$ flowing through the battery is given by the formula $I = \frac{E}{R_{eq} + r}$,where $E$ is the emf,$R_{eq}$ is the equivalent external resistance,and $r$ is the internal resistance.Since the two resistors $R_1 = 2 \Omega$ and $R_2 = 6 \Omega$ are connected in parallel,their equivalent resistance $R_{eq}$ is calculated as:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{6} = \frac{3 + 1}{6} = \frac{4}{6} = \frac{2}{3} \Omega^{-1}$.Therefore,$R_{eq} = \frac{3}{2} = 1.5 \Omega$.Now,substituting the values into the current formula:$I = \frac{2}{1.5 + 0.5} = \frac{2}{2.0} = 1 	ext{ A}$.Thus,the current flowing through the battery is $1 	ext{ A}$.

Two resistors of resistances $2 \Omega$ and $6 \Omega$ are connected in parallel. This combination is then connected to a battery of emf $2 \text{ V}$ and internal resistance $0.5 \Omega$. What is the current flowing through the battery?

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