The current I drawn from the 5 text{ V} sourc | vedclass

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(C) The circuit can be simplified by identifying series and parallel combinations. Looking at the circuit,the $10 \Omega$ and $20 \Omega$ resistors ar

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$0.5$

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(C) The circuit can be simplified by identifying series and parallel combinations. Looking at the circuit,the $10 \Omega$ and $20 \Omega$ resistors are in series,giving a total of $30 \Omega$. The $5 \Omega$ and $10 \Omega$ resistors are also in series,giving a total of $15 \Omega$. These two branches ($30 \Omega$ and $15 \Omega$) are in parallel with each other. The equivalent resistance $R_{eq}$ is given by: $R_{eq} = \frac{30 	imes 15}{30 + 15} = \frac{450}{45} = 10 \Omega$. Using Ohm's law,the current $I$ drawn from the $5 	ext{ V}$ source is: $I = \frac{V}{R_{eq}} = \frac{5}{10} = 0.5 	ext{ A}$.

The current $I$ drawn from the $5 \text{ V}$ source will be ............... $A$.

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