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(B) In steady state,the capacitor is fully charged,so no current flows through the branch containing the capacitor. Thus,the circuit simplifies to a s

Vedclass · Accepted Answer

$0.67$

Vedclass · Answer

(B) In steady state,the capacitor is fully charged,so no current flows through the branch containing the capacitor. Thus,the circuit simplifies to a single loop containing the two batteries and the two resistors of $3 \, \Omega$ and $9 \, \Omega$ in series. The net electromotive force $(EMF)$ in the loop is $E_{net} = 16.0 \, V - 8.0 \, V = 8.0 \, V$. The total resistance in the loop is $R_{total} = 3 \, \Omega + 9 \, \Omega = 12 \, \Omega$. Using Ohm's law,the current $I$ in the circuit is $I = \frac{E_{net}}{R_{total}} = \frac{8.0 \, V}{12 \, \Omega} = \frac{2}{3} \, A \approx 0.67 \, A$.

The circuit shown here has two batteries of $8.0 \, V$ and $16.0 \, V$,three resistors of $3 \, \Omega$,$9 \, \Omega$,and $9 \, \Omega$,and a capacitor of $5.0 \, \mu F$. How much is the current $I$ in the circuit in steady state? ................... $A$

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