Rankers Physics
Topic: Capacitors
Subtopic: Parallel Plate Capacitor

A 2μF capacitor is charged as shown in figure. The percentage of its stored energy dissiplated after the switch S is turned to position 2 is : Image related to
0%
20%
75%
80%

Solution:

To solve this, let's calculate the percentage of energy dissipated when the switch SS is turned to position 2.

Given:

  • C1=2μFC_1 = 2 \, \mu\text{F}
  • C2=8μFC_2 = 8 \, \mu\text{F}
  • Initial voltage across C1=VC_1 = V

Key Concept:

When two capacitors are connected, the redistribution of charge causes energy loss. The energy dissipated can be calculated using the direct formula:

Energy dissipated=12C1C2C1+C2(V1V2)2\text{Energy dissipated} = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2

Here:

  • V1V_1 is the initial voltage of C1C_1.
  • V2=0V_2 = 0 (initially, C2C_2 is uncharged).

Step-by-step Calculation:

  1. Common Voltage after Connection: The final voltage across both capacitors is:

    Vcommon=C1VC1+C2=2V2+8=V5.V_{\text{common}} = \frac{C_1 V}{C_1 + C_2} = \frac{2V}{2 + 8} = \frac{V}{5}.

  2. Initial Energy Stored in C1C_1:

    Einitial=12C1V2=12(2×106)V2=106V2J.E_{\text{initial}} = \frac{1}{2} C_1 V^2 = \frac{1}{2} (2 \times 10^{-6}) V^2 = 10^{-6} V^2 \, \text{J}.

  3. Final Energy Stored: The total energy stored after redistribution is:

    Efinal=12(C1+C2)Vcommon2=12(2+8)(V5)2.E_{\text{final}} = \frac{1}{2} (C_1 + C_2) V_{\text{common}}^2 = \frac{1}{2} (2 + 8) \left(\frac{V}{5}\right)^2.Substituting values:

    Efinal=12(10)(V225)=10V250=V25×106J.E_{\text{final}} = \frac{1}{2} (10) \left(\frac{V^2}{25}\right) = \frac{10 V^2}{50} = \frac{V^2}{5} \times 10^{-6} \, \text{J}.

  4. Energy Dissipated:

    Edissipated=EinitialEfinal=106V2106V25=45×106V2.E_{\text{dissipated}} = E_{\text{initial}} - E_{\text{final}} = 10^{-6} V^2 - \frac{10^{-6} V^2}{5} = \frac{4}{5} \times 10^{-6} V^2.

  5. Percentage of Energy Dissipated:

    %Dissipated=EdissipatedEinitial×100=45×106V2106V2×100=80%.\% \text{Dissipated} = \frac{E_{\text{dissipated}}}{E_{\text{initial}}} \times 100 = \frac{\frac{4}{5} \times 10^{-6} V^2}{10^{-6} V^2} \times 100 = 80\%.

Final Answer:

The percentage of energy dissipated is 80%.

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