Answer:
Capacitor [tex]C_{1}[/tex] would be damaged because the voltage across it, 75V, is far too greater than the required value, 25V.
Explanation:
A capacitor is an electronic device that can be used to store charges.
From the question, [tex]C_{1}[/tex] = 100uF, 25V and [tex]C_{2}[/tex] = 300uF, 35V, where voltage 'E' across the circuit is 100V.
Since the two capacitors are connected in series, the total capacitance, [tex]C_{T}[/tex], is;
[tex]C_{T}[/tex] = [tex]\frac{1}{C_{1} }[/tex] + [tex]\frac{1}{C_{2} }[/tex]
= [tex]\frac{C_{1} * C_{2} }{C_{1} + C_{2} }[/tex]
= [tex]\frac{100uF * 300uF}{100uF + 300uF}[/tex]
= 75uF
∴ [tex]C_{T}[/tex] = 75uF
Thus, the voltage drop across each capacitor are;
[tex]V_{c1}[/tex] = [tex]\frac{C_{T} }{C_{1} }[/tex] × E
= [tex]\frac{75uF}{100uF}[/tex] × 100
= 75V
[tex]V_{C2}[/tex] = [tex]\frac{C_{T} }{C_{2} }[/tex] × 100
= [tex]\frac{75uF}{300uF}[/tex] × 100
= 25V
The capacitor that would be damage is [tex]C_{1}[/tex] because the voltage across it, 75V, is more than the required value, 25V.
