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madreJ [45]
3 years ago
11

Two capacitors, C1 and C2, give an equivalent capacitance of 9.0 pF when connected in parallel and an equivalent capacitance of

2.0 pF when connected in series. What is the capacitance of each capacitor? C1 = _______ pF (smallest value) C2 = __________ pF (largest value)
Physics
1 answer:
Lilit [14]3 years ago
5 0

Answer:

C₁ = 3 pF

C₂ = 6 pF

Explanation:

Let the capacitance be C₁ and C₂.

For parallel combination

C₁ +C₂ = 9

For series combination

2=\frac{C_1\times C_2}{C_1+C_2}

2=\frac{C_1\times C_2}{9}

C₁ X C₂ = 18

( 9-C₂ )X C₂ = 18

C₂²- 9C₂ +18 = 0

C₂ = 6 or 3.

C₁ = 3 or 6.

C₁ = 3 pF

C₂ = 6 pF

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Read 2 more answers
Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters.
erma4kov [3.2K]

Answer:

The extent to which it would stretch  is \Delta L = 0.015 \ m

Explanation:

From the question we are told that

    The initial length is  L = 1.00m

     The area is  A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2

     The Young modulus of the steel is  Y = 2.0*10^{11} Pa

     The tension   is  T =1500 N

The Young modulus is mathematically represented as

       Y = \frac{\sigma}{e}

Where \sigma is the stress which is mathematically represented as

           \sigma = \frac{F}{A}  

Substituting values

            \sigma = \frac{1500}{0.500*10^{-6}}  

           \sigma = 3.0*10^9 N/m^2  

And  e is the strain which is mathematically represented as

            e = \frac{\Delta L}{L }

Where \Delta L The extension of the steel string

Substituting these into the equation above

             Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }

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          \Delta L = \frac{3.0*10^9  * 1}{2.0 *10^{11}}

         \Delta L = 0.015 \ m

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