Two capacitors of `10muF` and `20 muF` are connected in series with a `30 V` battery. The charge on the capacitors will be, respectively

Options

A. `100 muC, 200 muC`
B. `200 muC, 100 muC`
C. `100 muC, 100 muC`
D. `200 muC, 200 muC`

Explanation

To find the charge on the capacitors connected in series, we can follow these steps:

### Step 1: Calculate the equivalent capacitance (C_eq) of the capacitors in series.
For capacitors in series, the formula for equivalent capacitance is given by:

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

Where:
- \(C_1 = 10 \mu F\)
- \(C_2 = 20 \mu F\)

Substituting the values:

\[ \frac{1}{C_{eq}} = \frac{1}{10 \mu F} + \frac{1}{20 \mu F} \]

### Step 2: Find a common denominator and solve for C_eq.
The common denominator for \(10\) and \(20\) is \(20\):

\[ \frac{1}{C_{eq}} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} \]

Now, taking the reciprocal to find \(C_{eq}\):

\[ C_{eq} = \frac{20}{3} \mu F \approx 6.67 \mu F \]

### Step 3: Calculate the total charge (Q) stored in the equivalent capacitor using the formula:
The charge stored in a capacitor is given by:

\[ Q = C_{eq} \cdot V \]

Where \(V = 30 V\):

\[ Q = \left(\frac{20}{3} \mu F\right) \cdot 30 V \]

### Step 4: Convert capacitance to Farads and calculate Q.
First, convert \(C_{eq}\) to Farads:

\[ C_{eq} = \frac{20}{3} \times 10^{-6} F \]

Now, substituting into the equation for Q:

\[ Q = \left(\frac{20}{3} \times 10^{-6} F\right) \cdot 30 V = \frac{600}{3} \times 10^{-6} C = 200 \times 10^{-6} C = 200 \mu C \]

### Step 5: Determine the charge on each capacitor.
In a series connection, the charge on each capacitor is the same. Therefore, the charge on both capacitors is:

\[ Q_{10 \mu F} = Q_{20 \mu F} = 200 \mu C \]

### Final Answer:
The charge on both capacitors will be \(200 \mu C\) respectively.

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