Two, capacitors `A` and `B` are connected in series across a `100 V` supply and it is observed that the potential difference across them are `60 V` and `40 V`. A capacitor of `2 p.F` capacitance is now connected in parallel with `A` and the potential difference across B rises to `90 V`. Determine the capacitance of `A` and `B`

Explanation

To solve the problem, we need to determine the capacitances of capacitors A and B based on the given conditions. Let's break down the solution step by step.

### Step 1: Understand the Series Connection
When capacitors are connected in series, the total voltage across them is equal to the sum of the voltages across each capacitor. Given:
- Voltage across A (V_A) = 60 V
- Voltage across B (V_B) = 40 V
- Total Voltage (V_total) = 100 V

### Step 2: Use the Voltage Relationship
From the information provided, we can write:
\[ V_A + V_B = V_{total} \]
\[ 60 V + 40 V = 100 V \]
This confirms that the voltage distribution is correct.

### Step 3: Relate the Capacitances Using Voltage
In a series connection, the charge (Q) on each capacitor is the same. We can use the relationship:
\[ Q = C_A \cdot V_A = C_B \cdot V_B \]
This implies:
\[ \frac{C_A}{C_B} = \frac{V_B}{V_A} = \frac{40}{60} = \frac{2}{3} \]
Let \( C_B = 3x \) and \( C_A = 2x \).

### Step 4: Find the Total Capacitance
The total capacitance \( C_{total} \) for capacitors in series is given by:
\[ \frac{1}{C_{total}} = \frac{1}{C_A} + \frac{1}{C_B} \]
Substituting the values:
\[ \frac{1}{C_{total}} = \frac{1}{2x} + \frac{1}{3x} \]
Finding a common denominator:
\[ \frac{1}{C_{total}} = \frac{3 + 2}{6x} = \frac{5}{6x} \]
Thus:
\[ C_{total} = \frac{6x}{5} \]

### Step 5: Analyze the Second Condition
Now, a capacitor of 2 pF is connected in parallel with capacitor A. The new voltage across B becomes 90 V. The voltage across A must then be:
\[ V_A' = V_{total} - V_B' = 100 V - 90 V = 10 V \]

### Step 6: Update the Capacitance of A
The new capacitance \( C_A' \) when the 2 pF capacitor is added in parallel is:
\[ C_A' = C_A + 2 \text{ pF} \]

### Step 7: Relate the New Voltages
Using the new voltage across A (10 V) and the new voltage across B (90 V):
Since the charge remains the same:
\[ Q = C_A' \cdot V_A' = C_B \cdot V_B' \]
Substituting the values:
\[ (C_A + 2) \cdot 10 = C_B \cdot 90 \]

### Step 8: Substitute \( C_B \) and Solve
From our earlier relationship:
\[ C_B = 3x \]
So we have:
\[ (2x + 2) \cdot 10 = 3x \cdot 90 \]
Expanding and simplifying:
\[ 20x + 20 = 270x \]
\[ 250x = 20 \]
\[ x = \frac{20}{250} = \frac{2}{25} \text{ F} \]

### Step 9: Calculate Capacitances
Now substituting back to find \( C_A \) and \( C_B \):
\[ C_A = 2x = 2 \cdot \frac{2}{25} = \frac{4}{25} \text{ F} = 0.16 \text{ µF} \]
\[ C_B = 3x = 3 \cdot \frac{2}{25} = \frac{6}{25} \text{ F} = 0.24 \text{ µF} \]

### Final Answer
- Capacitance of A, \( C_A = 0.16 \text{ µF} \)
- Capacitance of B, \( C_B = 0.24 \text{ µF} \)

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