A 170-pF capacitor and a 680-pF capacitor are both charged to 1.70 kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate and negative plate to positive plate. (a) Find the resulting potential difference across each capacitor. V₁ kV kV 170 pF V₂ 680 pF = (b) Find the energy lost when the connections are made. µJ
A 170-pF capacitor and a 680-pF capacitor are both charged to 1.70 kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate and negative plate to positive plate. (a) Find the resulting potential difference across each capacitor. V₁ kV kV 170 pF V₂ 680 pF = (b) Find the energy lost when the connections are made. µJ
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Transcribed Image Text:**Capacitor Problem Overview**
**Initial Setup:**
- A 170-pF capacitor and a 680-pF capacitor are initially charged to 1.70 kV.
- Both capacitors are then disconnected from the voltage source.
- They are subsequently connected together with opposite plates (positive to negative and negative to positive).
**Tasks:**
**(a) Calculate the Potential Difference:**
- Determine the resulting potential difference across each capacitor after they are connected.
- \( V_{\text{170 pF}} = \) ____ kV
- \( V_{\text{680 pF}} = \) ____ kV
**(b) Calculate the Energy Lost:**
- Determine the energy lost during the connection.
- Energy lost = ____ µJ
**Instructions:**
- Approach the problem by considering the conservation of charge and energy.
**Note:**
- This activity helps explore concepts like capacitance, potential difference, and energy conservation in electrical circuits.
Expert Solution
Introduction:
When we apply voltage across conductor, it accumulates charge.
This charge is directly proportional to voltage. It is given as
Here is the capacitance.
We first find the charge on each capacitor and then total charge. We then find voltage across each capacitor when joined. We also find the loss of energy.
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