Two capacitors `C_1` and `C_2` are charged to 120V and 200V respectively. It is found that connecting them together the potential on each one can be made zero. Then

Options

A. `5C_(1)=3C_(2)`
B. `3C_(1)=5C_(2)`
C. `3C_(1)=5C_(2)=0`
D. `9C_(1)=4C_(2)`

Explanation

To solve the problem step by step, we need to analyze the situation where two capacitors, \( C_1 \) and \( C_2 \), are charged to different voltages and then connected in such a way that the potential across each capacitor becomes zero.

### Step 1: Understand the Initial Conditions
- Capacitor \( C_1 \) is charged to a voltage \( V_1 = 120 \, \text{V} \).
- Capacitor \( C_2 \) is charged to a voltage \( V_2 = 200 \, \text{V} \).

### Step 2: Determine the Charges on Each Capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \cdot V \]
Thus, the charges on the capacitors can be expressed as:
- \( Q_1 = C_1 \cdot V_1 = C_1 \cdot 120 \)
- \( Q_2 = C_2 \cdot V_2 = C_2 \cdot 200 \)

### Step 3: Set Up the Condition for Zero Potential
When the two capacitors are connected together, they can discharge or share their charges. For the potential across each capacitor to become zero, the charges must be equal in magnitude but opposite in sign. This means:
\[ Q_1 + Q_2 = 0 \]
or
\[ C_1 \cdot 120 + C_2 \cdot 200 = 0 \]

### Step 4: Rearranging the Equation
From the equation \( C_1 \cdot 120 + C_2 \cdot 200 = 0 \), we can express \( C_1 \) in terms of \( C_2 \):
\[ C_1 \cdot 120 = -C_2 \cdot 200 \]
Dividing both sides by 120 gives:
\[ C_1 = -\frac{200}{120} C_2 \]
This simplifies to:
\[ C_1 = -\frac{5}{3} C_2 \]

### Step 5: Considering the Absolute Values
Since we are dealing with magnitudes, we can drop the negative sign and express the relationship as:
\[ 3 C_1 = 5 C_2 \]

### Conclusion
The final relationship between the two capacitors is:
\[ 3 C_1 = 5 C_2 \]