7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?

Respuesta :

Answer:

q = C V    charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 *  10E-4  C per capacitor

N = Q / q = 1 / 1.1 * 10E-4  = 9091 capacitors

9.09 × 10³ capacitors must be connected in parallel.

How to calculate the number of capacitors connected in parallel?

Given C = 1.00μF = 1 × 10⁻⁶ F

          q = 1.00 C

          V = 110 V

The equivalent capacitance is given by

Ceq = q/V

where q = total charge on all the capacitors

             V = potential difference

For N number of identical capacitors in parallel,

Ceq = NC

Therefore,

NC = q/V

N = q/VC

Putting on the values in the above formula,

N = 1/ (110)(1 × 10⁻⁶)

   = 1 / 110 × 10⁻⁶

   = 9.09 × 10³

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