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Sagot :
voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]
[tex]c = \frac{c1 \times c2}{c1 + c2} [/tex]
[tex] = \frac{2 \times 4}{2 + 4} [/tex]
=1.33μf
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.
Q=CV
given,V=8v
[tex] = 1.33 \times 10 {}^{ - 6} \times 8[/tex]
[tex] = 10.64 \times 10 {}^{ - 6} [/tex]
charge on 2.0μf capacitor is
[tex] \frac{Qeq}{2 \times 10 {}^{ - 6} } [/tex]
[tex] = \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} } [/tex]
=5.32v
learn more about series capacitance from here: https://brainly.com/question/28166078
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