To achieve a maximum effective capacitance using six identical capacitors each of capacitance \( C \) and voltage rating of \( 12 , \text{V} \), with the requirement that the resulting capacitor can withstand \( 36 , \text{V} \), we need to connect the capacitors appropriately.
Step 1: Voltage Configuration
Given that each capacitor can only withstand \( 12 , \text{V} \), to achieve an effective capacitor that can handle \( 36 , \text{V} \), we must connect some of the capacitors in series.
To reach \( 36 , \text{V} \), we can use 3 capacitors in series since:
\[ V_{\text{total}} = V_1 + V_2 + V_3 = 12, \text{V} + 12, \text{V} + 12, \text{V} = 36, \text{V} \]
This configuration allows us to build a sub-capacitor that can withstand \( 36 , \text{V} \).
Step 2: Capacity Calculation for Series Configuration
When capacitors are connected in series, the total capacitance \( C_s \) is given by:
\[ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \]
Thus, the equivalent capacitance for three capacitors in series is:
\[ C_s = \frac{C}{3} \]
Step 3: Remaining Capacitors for Parallel Configuration
After using up 3 capacitors in series, we still have 3 capacitors left. These remaining capacitors can be connected in parallel to the series combination.
The capacitance of the parallel capacitors \( C_p \) would simply add together:
\[ C_p = C + C + C = 3C \]
Step 4: Total Effective Capacitance Calculation
Now, we have \( C_s \) (the series combination) and \( C_p \) (the parallel combination). The total effective capacitance \( C_{\text{total}} \) is then:
\[ C_{\text{total}} = C_s + C_p = \frac{C}{3} + 3C \]
To combine these fractions:
First, convert \( 3C \) into a fraction with the same denominator:
\[ C_p = 3C = \frac{9C}{3} \]
Now add the two:
\[ C_{\text{total}} = \frac{C}{3} + \frac{9C}{3} = \frac{10C}{3} \]
Final Answer
The maximum effective capacitance that can be achieved while ensuring that the configuration can withstand \( 36, \text{V} \) is:
\[ \boxed{\frac{10C}{3}} \]
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