1. **Problem Statement:** We have a complex capacitor network with capacitors $C_1=1\,\mu F$, $C_2=2\,\mu F$, $C_3=3\,\mu F$, $C_4=4\,\mu F$, $C_5=5\,\mu F$, and $C_6=6\,\mu F$ connected to a $48\,V$ battery. We need to find the charge ($Q_i$), voltage ($V_i$), and potential energy ($PE_i$) for each capacitor, as well as the total charge and total capacitance. 2. **Key Formulas and Rules:** - Charge on capacitor: $Q = CV$ - Voltage across capacitor: $V = \frac{Q}{C}$ - Potential energy stored: $PE = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$ - Capacitors in series: $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}$ - Capacitors in parallel: $C_{eq} = \sum C_i$ 3. **Analyze the circuit:** - Capacitors $C_2$ and $C_3$ form a triangle on the left bottom side, likely in series or parallel. - Capacitors $C_5$ and $C_6$ form the rightmost vertical side, likely in series or parallel. - Capacitor $C_4$ connects between $C_1$ and $C_6$. - $C_1$ is at the top side horizontally. 4. **Simplify the network step-by-step:** - Combine $C_2$ and $C_3$ in series: $$\frac{1}{C_{23}} = \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \Rightarrow C_{23} = \frac{6}{5} = 1.2\,\mu F$$ - Combine $C_5$ and $C_6$ in series: $$\frac{1}{C_{56}} = \frac{1}{C_5} + \frac{1}{C_6} = \frac{1}{5} + \frac{1}{6} = \frac{6}{30} + \frac{5}{30} = \frac{11}{30} \Rightarrow C_{56} = \frac{30}{11} \approx 2.727\,\mu F$$ - Now, $C_4$ is connected between $C_1$ and $C_6$, so $C_4$ is in parallel with $C_{56}$: $$C_{456} = C_4 + C_{56} = 4 + 2.727 = 6.727\,\mu F$$ - Finally, $C_1$, $C_{23}$, and $C_{456}$ are connected in series: $$\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_{23}} + \frac{1}{C_{456}} = 1 + \frac{1}{1.2} + \frac{1}{6.727} = 1 + 0.8333 + 0.1486 = 1.9819$$ $$C_{total} = \frac{1}{1.9819} \approx 0.5046\,\mu F$$ 5. **Calculate total charge:** $$Q_{total} = C_{total} \times V = 0.5046 \times 48 = 24.22\,\mu C$$ 6. **Calculate voltages across each series branch:** - Voltage across $C_1$: $$V_1 = \frac{Q_{total}}{C_1} = \frac{24.22}{1} = 24.22\,V$$ - Voltage across $C_{23}$: $$V_{23} = \frac{Q_{total}}{C_{23}} = \frac{24.22}{1.2} = 20.18\,V$$ - Voltage across $C_{456}$: $$V_{456} = \frac{Q_{total}}{C_{456}} = \frac{24.22}{6.727} = 3.60\,V$$ 7. **Calculate charges and voltages on $C_2$ and $C_3$ (series capacitors):** - Charge is same on series capacitors: $Q_2 = Q_3 = Q_{23} = 24.22\,\mu C$ - Voltages: $$V_2 = \frac{Q_2}{C_2} = \frac{24.22}{2} = 12.11\,V$$ $$V_3 = \frac{Q_3}{C_3} = \frac{24.22}{3} = 8.07\,V$$ 8. **Calculate charges and voltages on $C_5$ and $C_6$ (series capacitors):** - Charge is same: $Q_5 = Q_6 = Q_{56}$ - Voltage across $C_4$ is $V_4 = V_{456} - V_{56}$, but since $C_4$ is in parallel with $C_{56}$, voltage across $C_4$ equals voltage across $C_{56}$, so: $$V_4 = V_{456} = 3.60\,V$$ - Voltage across $C_5$ and $C_6$ sum to $V_{56} = V_4 = 3.60\,V$ - Using voltage division: $$V_5 = V_{56} \times \frac{C_6}{C_5 + C_6} = 3.60 \times \frac{6}{11} = 1.96\,V$$ $$V_6 = V_{56} \times \frac{C_5}{C_5 + C_6} = 3.60 \times \frac{5}{11} = 1.64\,V$$ - Charges: $$Q_5 = C_5 \times V_5 = 5 \times 1.96 = 9.8\,\mu C$$ $$Q_6 = C_6 \times V_6 = 6 \times 1.64 = 9.84\,\mu C$$ 9. **Calculate potential energies:** - For each capacitor: $$PE_i = \frac{1}{2} C_i V_i^2$$ - Calculate each: $$PE_1 = 0.5 \times 1 \times 24.22^2 = 293.3\,\mu J$$ $$PE_2 = 0.5 \times 2 \times 12.11^2 = 146.7\,\mu J$$ $$PE_3 = 0.5 \times 3 \times 8.07^2 = 97.7\,\mu J$$ $$PE_4 = 0.5 \times 4 \times 3.60^2 = 25.9\,\mu J$$ $$PE_5 = 0.5 \times 5 \times 1.96^2 = 9.6\,\mu J$$ $$PE_6 = 0.5 \times 6 \times 1.64^2 = 8.1\,\mu J$$ 10. **Summary:** - Charges: $Q_1=24.22$, $Q_2=24.22$, $Q_3=24.22$, $Q_4=24.22$, $Q_5=9.8$, $Q_6=9.84$ (all in $\mu C$) - Voltages: $V_1=24.22$, $V_2=12.11$, $V_3=8.07$, $V_4=3.60$, $V_5=1.96$, $V_6=1.64$ (all in V) - Potential energies: $PE_1=293.3$, $PE_2=146.7$, $PE_3=97.7$, $PE_4=25.9$, $PE_5=9.6$, $PE_6=8.1$ (all in $\mu J$) - Total charge: $Q_{total} = 24.22\,\mu C$ - Total capacitance: $C_{total} = 0.5046\,\mu F$