A battery of 9 , V is connected in series wit | vedclass

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(C) In a series circuit,the current remains the same through all components. According to Ohm's law,the current $I$ is given by $I = \frac{V}{R_{eq}}$

Vedclass · Accepted Answer

$0.67$

Vedclass · Answer

(C) In a series circuit,the current remains the same through all components. According to Ohm's law,the current $I$ is given by $I = \frac{V}{R_{eq}}$.The equivalent resistance $R_{eq}$ for resistors connected in series is the sum of individual resistances:$R_{eq} = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \, \Omega$.Given the potential difference $V = 9 \, V$,the current is:$I = \frac{9}{13.4} \approx 0.671 \, A$.Thus,the current flowing through the $12 \, \Omega$ resistor is approximately $0.67 \, A$.

$A$ battery of $9 \, V$ is connected in series with resistors of $0.2 \, \Omega$,$0.3 \, \Omega$,$0.4 \, \Omega$,$0.5 \, \Omega$,and $12 \, \Omega$ respectively. How much current (in $A$) would flow through the $12 \, \Omega$ resistor?

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