Short Answer 
The equivalent capacitance for Figure 1 is approximately 2.09 ¬¨¬µF, while for Figure 2, it’s around 1.73 ¬¨¬µF. The calculations include combining capacitors in parallel and series as per their configurations.
Step 1: Calculate Equivalent Capacitance for Figure 1
To find the equivalent capacitance of the capacitors in Figure 1, first calculate the combined capacitance of capacitors C3 and C4 which are in parallel. Use the formula:
- C34 = C3 + C4 = 3.00 µF + 5.00 µF = 8.00 µF
Next, find the equivalent capacitance of C2 and C34 using the series formula:
- 1/C234 = 1/C2 + 1/C34
- C234 = 4.40 µF (approx.)
Finally, combine C1 and C234 to determine the total capacitance:
- 1/C_total = 1/C1 + 1/C234
- C_total = 2.09 µF (approx.)
Step 2: Calculate Equivalent Capacitance for Figure 2
In Figure 2, find the equivalent capacitance of capacitors C3, C5, and C6 which are connected in parallel. Use the formula:
- C356 = C3 + C5 + C6 = 3.00 µF + 6.00 µF + 3.00 µF = 12.00 µF
Then, calculate the equivalent capacitance of C2 and C356 in series:
- 1/C2356 = 1/C2 + 1/C356
- C2356 = 4.89 µF (approx.)
Step 3: Determine Total Capacitance for Figure 2
To find the overall equivalent capacitance for the entire network in Figure 2, combine C1 with C2356, again using the series formula:
- 1/C_total = 1/C1 + 1/C2356
- C_total = 1.73 µF (approx.)
Therefore, the final values for the equivalent capacitances are approximately 2.09 µF for Figure 1 and 1.73 µF for Figure 2.