Question : (a)Under steady-state conditions, find the unknown currents I...

First note that because the capacitor represents an open circuit, there is no current between g and b along path g h a b under steady-state conditions. Therefore, when the charges associated with I_{1}  reach point g, they all go through the 8.00 \mathrm{~V}  battery to point b; hence, I_{g b}=I_{1} . Labeling the currents as shown in Figure 28.15 and applying Equation 28.9 to junction c, we obtain (1) I_{1}+I_{2}=I_{3}

Equation 28.10 applied to loops defcd and cgbc, traversed clockwise, gives

(2) d e f d \quad 4.00 \mathrm{~V}-(8.00 \Omega) I_{2}-(5.00 \Omega) I_{3}=0

(3) \operatorname{cgbc} (8.00 \Omega) I_{2}-(5.00 \Omega) I_{1}+8.00 \mathrm{~V}=0

From Equation (1) we see that I_{1}=I_{3}-I_{2},  which, when substituted into Equation (3), gives (4)(8.00 \Omega) I_{2}-(5.00 \Omega) I_{3}+8.00 \mathrm{~V}=0

Subtracting Equation (4) from Equation (2), we eliminate I_{3}  and find that

Because our value for I_{2}  is negative, we conclude that the direction of I_{2}  is from c to f through the 3.00-\Omega  resistor. Despite this interpretation of the direction, however, we must continue to use this negative value for I_{2}  in subsequent calculations because our equations were established with our original choice of direction. Using I_{2}=-0.364  \mathrm{~A}  in Equations ( 3 ) and (1) gives I_{1}=1.38 \mathrm{~A} \quad I_{3}=1.02 \mathrm{~A}

(b) What is the charge on the capacitor?

Solution We can apply Kirch hoff’s loop rule to loop bghab (or any other loop that contains the capacitor) to find the potential difference \Delta V_{\text {cap }} across the capacitor. We enter this potential difference in the equation without reference to a sign convention because the charge on the capacitor depends only on the magnitude of the potential difference. Moving clockwise around this loop, we obtain

Because Q=C \Delta V_{\text {cap }}  (see Eq. 26.1 ), the charge on the capacitor is