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 pF` 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 information. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
We have two capacitors, A and B, connected in series across a 100 V supply. The potential difference across A is 60 V, and across B is 40 V.

### Step 2: Use the Relationship of Capacitors in Series
For capacitors in series, the charge (Q) on each capacitor is the same. The relationship between capacitance (C), charge (Q), and voltage (V) is given by:
\[ Q = C \cdot V \]

From this, we can express the capacitances in terms of the voltages:
\[ C_A = \frac{Q}{V_A} \quad \text{and} \quad C_B = \frac{Q}{V_B} \]

Since \( Q \) is the same for both capacitors, we can write:
\[ \frac{C_A}{C_B} = \frac{V_B}{V_A} \]

Substituting the values we have:
\[ \frac{C_A}{C_B} = \frac{40}{60} = \frac{2}{3} \]

### Step 3: Express C_A in Terms of C_B
From the above relationship, we can express \( C_A \) in terms of \( C_B \):
\[ C_A = \frac{2}{3} C_B \]
Let’s call this Equation (1).

### Step 4: Analyze the New Configuration
Now, a capacitor of 2 pF is connected in parallel with capacitor A. The new capacitance of A becomes:
\[ C_A' = C_A + 2 \, \text{pF} \]

The potential difference across B rises to 90 V, and we need to find the new potential difference across A:
Since the total voltage remains 100 V, we can find the new voltage across A:
\[ V_A' = 100 \, \text{V} - V_B' = 100 \, \text{V} - 90 \, \text{V} = 10 \, \text{V} \]

### Step 5: Set Up the New Relationship
Using the same relationship for capacitors in series again:
\[ \frac{C_A'}{C_B} = \frac{V_B'}{V_A'} \]

Substituting the known values:
\[ \frac{C_A + 2}{C_B} = \frac{90}{10} = 9 \]

### Step 6: Substitute C_A from Equation (1)
Now substitute \( C_A \) from Equation (1):
\[ \frac{\frac{2}{3} C_B + 2}{C_B} = 9 \]

### Step 7: Solve for C_B
Cross-multiplying gives:
\[ \frac{2}{3} C_B + 2 = 9 C_B \]
Rearranging terms:
\[ 2 = 9 C_B - \frac{2}{3} C_B \]
\[ 2 = \left(9 - \frac{2}{3}\right) C_B \]
To combine the terms:
\[ 9 = \frac{27}{3} \]
So:
\[ 9 - \frac{2}{3} = \frac{27}{3} - \frac{2}{3} = \frac{25}{3} \]
Thus:
\[ 2 = \frac{25}{3} C_B \]
Multiplying both sides by \( \frac{3}{25} \):
\[ C_B = \frac{2 \cdot 3}{25} = \frac{6}{25} \, \text{pF} \]

### Step 8: Calculate C_A
Now substitute \( C_B \) back into Equation (1):
\[ C_A = \frac{2}{3} C_B = \frac{2}{3} \cdot \frac{6}{25} = \frac{12}{75} = \frac{4}{25} \, \text{pF} \]

### Final Answer
Thus, the capacitances are:
- \( C_A = \frac{4}{25} \, \text{pF} \) or \( 0.16 \, \text{pF} \)
- \( C_B = \frac{6}{25} \, \text{pF} \) or \( 0.24 \, \text{pF} \)