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When two capacitors are connected in ser...

When two capacitors are connected in series, the equivalent capacitance is `1.2muF`. When they are connected in parallel, the equivalent capacitance is `5muF`. The capacitances of the capacitors are

A

`3muFand2muF`

B

`4muFand1muF`

C

`2.5muFand2.5muF`

D

`3.5muFand1.5muF`

Text Solution

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The correct Answer is:
To find the capacitances of the two capacitors \( C_1 \) and \( C_2 \) based on the given conditions, we can follow these steps: ### Step 1: Understand the formulas for capacitors in series and parallel When capacitors are connected in series, the equivalent capacitance \( C_s \) is given by: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \] When capacitors are connected in parallel, the equivalent capacitance \( C_p \) is given by: \[ C_p = C_1 + C_2 \] ### Step 2: Set up the equations based on the problem statement From the problem, we know: - \( C_s = 1.2 \, \mu F \) - \( C_p = 5 \, \mu F \) Using the series formula, we can rearrange it to: \[ C_1 C_2 = C_s (C_1 + C_2) \] Substituting \( C_s = 1.2 \, \mu F \): \[ C_1 C_2 = 1.2 (C_1 + C_2) \] Using the parallel formula: \[ C_1 + C_2 = 5 \, \mu F \] ### Step 3: Substitute \( C_1 + C_2 \) into the series equation Let \( C_1 + C_2 = 5 \, \mu F \). We can substitute this into our earlier equation: \[ C_1 C_2 = 1.2 \times 5 \] \[ C_1 C_2 = 6 \, \mu F^2 \] ### Step 4: Set up a system of equations Now we have two equations: 1. \( C_1 + C_2 = 5 \) 2. \( C_1 C_2 = 6 \) ### Step 5: Solve the equations Let's denote \( C_1 \) and \( C_2 \) as the roots of the quadratic equation: \[ x^2 - (C_1 + C_2)x + C_1 C_2 = 0 \] Substituting the known values: \[ x^2 - 5x + 6 = 0 \] ### Step 6: Factor the quadratic equation Factoring the equation: \[ (x - 2)(x - 3) = 0 \] Thus, the solutions are: \[ x = 2 \quad \text{or} \quad x = 3 \] ### Step 7: Identify the capacitances So, we have: - \( C_1 = 2 \, \mu F \) - \( C_2 = 3 \, \mu F \) ### Conclusion The capacitances of the two capacitors are \( C_1 = 2 \, \mu F \) and \( C_2 = 3 \, \mu F \).

To find the capacitances of the two capacitors \( C_1 \) and \( C_2 \) based on the given conditions, we can follow these steps: ### Step 1: Understand the formulas for capacitors in series and parallel When capacitors are connected in series, the equivalent capacitance \( C_s \) is given by: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \] When capacitors are connected in parallel, the equivalent capacitance \( C_p \) is given by: ...
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Knowledge Check

  • When two capacitors are connected in series, the equivalent capacitance is (15)/(4) muF . When they are connected in parallel, the equivalent capacitance is 16muF . The individual capacitance are

    A
    `5 muF, 11 muF`
    B
    `6 muF, 10 muF`
    C
    `4 muF, 12 muF`
    D
    `8 muF, 8 muF`
  • When two capacitances are connected in series, the equivalent capacitance is 2.4muF and when they connected in parallel, it is 10muF . The individual capacitances are

    A
    `6muFand4muF`
    B
    `5muFand5muF`
    C
    `7muFand3muF`
    D
    `8muFand2muF`
  • Two identical capacitors when joined in series have an effective capacitance of 3 muF . When connected in parallel, the effective capacitance becomes 12 muF . What is the capacitance of each capacitor ?

    A
    `6 mu F`
    B
    ` 3 mu F `
    C
    `12 mu F `
    D
    `9 mu F`
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