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(C) When the switch $S$ is open,the $10\,\Omega$ and $20\,\Omega$ resistors are in series. The total resistance $R_{eq} = 10\,\Omega + 20\,\Omega = 30

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$\frac{1}{10} 	ext{ A}, \frac{3}{10} 	ext{ A}$

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(C) When the switch $S$ is open,the $10\,\Omega$ and $20\,\Omega$ resistors are in series. The total resistance $R_{eq} = 10\,\Omega + 20\,\Omega = 30\,\Omega$. The current $I = \frac{V}{R_{eq}} = \frac{3 	ext{ V}}{30\,\Omega} = \frac{1}{10} 	ext{ A}$.$(b)$ When the switch $S$ is closed,the $20\,\Omega$ resistor is short-circuited. The circuit now only contains the $10\,\Omega$ resistor in series with the battery. The total resistance $R_{eq} = 10\,\Omega$. The current $I = \frac{V}{R_{eq}} = \frac{3 	ext{ V}}{10\,\Omega} = \frac{3}{10} 	ext{ A}$.Thus,the currents are $\frac{1}{10} 	ext{ A}$ and $\frac{3}{10} 	ext{ A}$ respectively.

Consider the circuit shown in the figure. Find the current through the $10\,\Omega$ resistor when the switch $S$ is $(a)$ open $(b)$ closed.

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