Three capacitors of `3muF` each are connected in series. This combination is connected in series to another combination of three capacitors of `1muF` each in parallel. Find the total capacitance.

Answer

To solve the problem, we need to find the total capacitance of the given capacitor configurations step by step.

### Step 1: Calculate the equivalent capacitance of the first combination (3 capacitors of 3μF in series)

When capacitors are connected in series, the formula for the equivalent capacitance (C_eq) is given by:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

For three capacitors of 3μF each:

\[ \frac{1}{C_{eq1}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \]

Calculating this gives:

\[ \frac{1}{C_{eq1}} = \frac{3}{3} = 1 \implies C_{eq1} = 1μF \]

### Step 2: Calculate the equivalent capacitance of the second combination (3 capacitors of 1μF in parallel)

When capacitors are connected in parallel, the equivalent capacitance is simply the sum of their capacitances:

\[ C_{eq2} = C_1 + C_2 + C_3 \]

For three capacitors of 1μF each:

\[ C_{eq2} = 1 + 1 + 1 = 3μF \]

### Step 3: Combine the two equivalent capacitances in series

Now, we have the two equivalent capacitances: \(C_{eq1} = 1μF\) and \(C_{eq2} = 3μF\). These two combinations are connected in series, so we will use the series formula again:

\[ \frac{1}{C_{total}} = \frac{1}{C_{eq1}} + \frac{1}{C_{eq2}} \]

Substituting the values:

\[ \frac{1}{C_{total}} = \frac{1}{1} + \frac{1}{3} \]

Calculating this gives:

\[ \frac{1}{C_{total}} = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \]

Thus,

\[ C_{total} = \frac{3}{4}μF = 0.75μF \]

### Final Answer

The total capacitance of the given configuration is:

\[ C_{total} = 0.75μF \]

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