Respuesta :
Explanation:
The given data is as follows.
[tex]C_{1} = 1.50 \times 10^{-6} F[/tex]
[tex]C_{1} = 3.50 \times 10^{-6} F[/tex]
Voltage = 2.50 V
Hence, calculate the equivalence capacitor as follows.
[tex]\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}[/tex]
[tex]\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}[/tex]
= [tex]0.945 \times 10^{-6} F[/tex]
C = [tex]1.06 \times 10^{-6} F[/tex]
Now, we will calculate the charge across each capacitance as follows.
Q = CV
= [tex]1.06 \times 10^{-6} F \times 2.50 V[/tex]
= [tex]2.65 \times 10^{-6} C[/tex]
= [tex]2.65 \mu C[/tex]
Thus, we can conclude that [tex]2.65 \mu C[/tex] is the charge stored on each given capacitor.
