JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Two capacitors \(C_1\) and \(C_2\) are charged to \(120\ V\) and \(200\ V\) respectively. It is found that connecting them together the potential on each one can be made zero. Then
- A \(9C_1=4C_2\)
- B \(5C_1=3C_2\)
- C \(3C_1=5C_2\)
- D \(3C_1+5C_2=0\)
Answer & Solution
Correct Answer
(C) \(3C_1=5C_2\)
Step-by-step Solution
Detailed explanation
For potential to be made zero, after connection \(120 \mathrm{C}_{1}=200 \mathrm{C}_{2} \quad\left[\because C=\frac{q}{v}\right]\) \(\Rightarrow \quad 3 C_{1}=5 C_{2}\)
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