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

Explanation

To solve the problem of finding the charge on two capacitors connected in series with a 30 V battery, we will follow these steps:

### Step 1: Identify the capacitance values
The given capacitance values are:
- \( C_1 = 10 \, \mu F \)
- \( C_2 = 20 \, \mu F \)

### Step 2: Calculate the equivalent capacitance for capacitors in series
For capacitors in series, the formula for equivalent capacitance \( C \) is given by:

\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]

Substituting the values:

\[ \frac{1}{C} = \frac{1}{10 \, \mu F} + \frac{1}{20 \, \mu F} \]

### Step 3: Calculate the individual fractions
Calculating the individual fractions:

\[ \frac{1}{C} = \frac{2}{20 \, \mu F} + \frac{1}{20 \, \mu F} = \frac{2 + 1}{20 \, \mu F} = \frac{3}{20 \, \mu F} \]

### Step 4: Find the equivalent capacitance
Now, we can find \( C \):

\[ C = \frac{20 \, \mu F}{3} \]

### Step 5: Calculate the total charge using the battery voltage
The total charge \( Q \) stored in the equivalent capacitor when connected to a voltage \( V \) is given by:

\[ Q = C \times V \]

Substituting the values:

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

### Step 6: Perform the multiplication
Calculating the charge:

\[ Q = \frac{20 \times 30}{3} \, \mu C = \frac{600}{3} \, \mu C = 200 \, \mu C \]

### Step 7: Conclusion
Since both capacitors are in series, they will have the same charge. Therefore, the charge on both capacitors is:

\[ Q_1 = Q_2 = 200 \, \mu C \]

### Final Answer
The charge on the capacitors will be, respectively, \( 200 \, \mu C \) and \( 200 \, \mu C \).

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