Question

You are given six identical capacitors of capacitance C that can each withstand a potential difference of 12V across them. These must be interconnected in such a way as to fabricate a capacitor capable of working with a potential difference of 36V across it. What is the maximum effective capacitance that can be achieved?

Answer 1

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:

\[
\boxed{\frac{10C}{3}}
\]