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Half the potential difference of the the1-µF
A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.
The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.
Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.
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Half the potential difference of the 1μf capacitor.
Charge: Q = CV where C is the capacitance in Farads, V is the voltage across the capacitor in Volts and Q is the charge measured in coulombs (C).
Energy stored: W = ½ QV = ½ CV×V
where W is the energy measured in Joules.
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
When the capacitors are connected in the form of series combination, the capacitance in total is less than the individual capacitances of the series capacitors. If one, two or number of capacitors are connected in the series form, the overall effect is the single or equivalent capacitor which has the total sum of the spacings between the plates of the individual capacitors. The increase in the plate spacing results in the decreased capacitance, with all the other factors remaining unchanged.
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