The current drawn from a 5 V source in the g | vedclass

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(B) First,simplify the circuit. The $5 \ \Omega$ and $10 \ \Omega$ resistors are in series,giving $R_1 = 5 + 10 = 15 \ \Omega$.Next,the $10 \ \Omega$

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$0.5$

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(B) First,simplify the circuit. The $5 \ \Omega$ and $10 \ \Omega$ resistors are in series,giving $R_1 = 5 + 10 = 15 \ \Omega$.Next,the $10 \ \Omega$ and $20 \ \Omega$ resistors are in series,giving $R_2 = 10 + 20 = 30 \ \Omega$.These two branches ($R_1$ and $R_2$) are in parallel. The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{15} + \frac{1}{30} = \frac{2+1}{30} = \frac{3}{30} = \frac{1}{10}$So,$R_{eq} = 10 \ \Omega$.The current $I$ is given by Ohm's law: $I = \frac{V}{R_{eq}} = \frac{5 \ V}{10 \ \Omega} = 0.5 \ A$.

The current drawn from a $5 \ V$ source in the given circuit is .......... $A$.

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