The currents are steady in the circuit, find \( 11,12,13,14,15 \), and the charge on the capacitor. The currents are steady in the circuit of Figure 4, Find \( I_{1}, I_{2}, I_{3}, I_{4}, I_{5} \), and the charge on the capacitor.
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The currents are steady in the circuit of $\underline{\text { Fig. } 29-4 .}$ Find $I_{1}, I_{2}, I_{3}$, $I_{4}, I_{5}$, and the charge on the capacitor. The capacitor passes no current when charged, and so $I_{5}=0$. Consider loop acba. The loop rule leads to $$ -8.0+4.0 I_{2}=0 \quad \text { or } \quad I_{2}=2.0 \mathrm{~A} $$ Using loop adeca gives $$ -3.0 I_{1}-9.0+8.0=0 \quad \text { or } \quad I_{1}=-0.33 \mathrm{~A} $$ Applying the node rule at point- $c$ results in $$ I_{1}+I_{5}+I_{2}=I_{3} \quad \text { or } \quad I_{3}=1.67 \mathrm{~A}=1.7 \mathrm{~A} $$ and at point- $a$, it yields $$ I_{3}=I_{4}+I_{2} \quad \text { or } \quad I_{4}=-0.33 \mathrm{~A} $$ (We should have realized this at once, because $I_{5}=0$ and so $I_{4}=$ $I_{1} .$. To find the charge on the capacitor, we need the voltage $V_{f g}$ across it. Put in all the signs on the resistors, batteries, and capacitor. Applying the loop rule to loop dfgced gives $$ -2.0 I_{5}+V_{f g}-7.0+9.0+3.0 I_{1}=0 \quad \text { or } \quad 0+V_{f g}-7.0+9.0-1.0=0 $$ from which $V_{f g}=-1.0 \mathrm{~V} .$ The minus sign tells us that plate $g$ is negative. The capacitor's charge is $$ Q=C V=(5.0 \mu \mathrm{F})(1.0 \mathrm{~V})=5.0 \mu \mathrm{C} $$
The currents in the circuit are steady. Find $I_{1}, I_{2}, I_{3}$, and the charge on the capacitor. When a capacitor has a constant charge, as it does here, the current flowing to it is zero. Therefore, $I_{2}=0$, and the circuit behaves just as though the center wire were missing. With the center wire missing, the remaining circuit is simply 12 $\Omega$ connected across a 15 -V battery. Therefore, $$ I_{1}=\frac{\varepsilon}{R}=\frac{15 \mathrm{~V}}{12 \Omega}=1.25 \mathrm{~A} $$ In addition, because $I_{2}=0$, we have $I_{3}=I_{1}=1.3 \mathrm{~A}$. To find the charge on the capacitor, first find the voltage difference between points- $a$ and $-b$. Start at $a$ and go around the upper path. Voltage change from $a$ to $b=-(5.0 \Omega) I_{3}+6.0 \mathrm{~V}+(3.0 \Omega) I_{2}$ $$ =-(5.0 \Omega)(1.25 \mathrm{~A})+6.0 \mathrm{~V}+(3.0 \Omega)(0)=-0.25 \mathrm{~V} $$ Therefore, $b$ is at the lower potential and the capacitor plate at $b$ is negative. To find the charge on the capacitor, $$ Q=C V_{a b}=\left(2 \times 10^{-6} \mathrm{~F}\right)(0.25 \mathrm{~V})=0.5 \mu \mathrm{C} $$
A Circuit is wired up as shown below. The capacitor is initially uncharged. A long time after the switch has been closed, what is the voltage across the capacitor? V = 10V R = 10Ω C = 0.0010 F R = 10Ω
Mahipal K.
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