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Two capacitors connected in series C1: 100uF, 25V and C2: 300uF, 35V, both are supplied by E= 100V. Verify which capacitor will be damaged

User Drbunsen
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1 Answer

6 votes

Answer:

Capacitor
C_(1) would be damaged because the voltage across it, 75V, is far too greater than the required value, 25V.

Step-by-step explanation:

A capacitor is an electronic device that can be used to store charges.

From the question,
C_(1) = 100uF, 25V and
C_(2) = 300uF, 35V, where voltage 'E' across the circuit is 100V.

Since the two capacitors are connected in series, the total capacitance,
C_(T), is;


C_(T) =
(1)/(C_(1) ) +
(1)/(C_(2) )

=
(C_(1) * C_(2) )/(C_(1) + C_(2) )

=
(100uF * 300uF)/(100uF + 300uF)

= 75uF


C_(T) = 75uF

Thus, the voltage drop across each capacitor are;


V_(c1) =
(C_(T) )/(C_(1) ) × E

=
(75uF)/(100uF) × 100

= 75V


V_(C2) =
(C_(T) )/(C_(2) ) × 100

=
(75uF)/(300uF) × 100

= 25V

The capacitor that would be damage is
C_(1) because the voltage across it, 75V, is more than the required value, 25V.

User Urho
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7.7k points