In the given circuit,the current flowing thro | vedclass

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(C) The circuit consists of two batteries connected in series with a resistor of $38 \ \Omega$. Looking at the polarities,the $10 \ V$ battery and the

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$5$

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(C) The circuit consists of two batteries connected in series with a resistor of $38 \ \Omega$. Looking at the polarities,the $10 \ V$ battery and the $200 \ V$ battery are connected in opposition (positive terminals facing each other). Therefore,the net electromotive force $(EMF)$ in the circuit is $E_{net} = 200 \ V - 10 \ V = 190 \ V$. The total resistance in the circuit is $R = 38 \ \Omega$. According to Ohm's law,the current $I$ flowing through the circuit is given by $I = \frac{E_{net}}{R}$. Substituting the values,we get $I = \frac{190 \ V}{38 \ \Omega} = 5 \ A$. Thus,the current flowing through the circuit is $5 \ A$.

In the given circuit,the current flowing through the circuit is: (in $A$)

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