Answer:
a) [tex]V_c=0[/tex]
b) [tex]V_R=145V[/tex]
c) [tex]Q_c=0[/tex]
d) [tex]I=\frac{1}{60}A[/tex]
Explanation:
From the question we are told that:
Capacitor [tex]C=4.60[/tex]
Resistor [tex]R=7.50[/tex]
Source emf [tex]E=125V[/tex]
a)
Generally The voltage drop across the capacitor is
V_c=0
b)
Generally the equation for Voltage drop is mathematically given by
[tex]V=IR[/tex]
[tex]V=\frac{E}{R}*R[/tex]
[tex]V_R=145V[/tex]
c
Generally The Charge across the capacitor is
[tex]Q_c=0[/tex]
d)
Generally the equation for Current is mathematically given by
[tex]I=\frac{V}{R}[/tex]
[tex]I=\frac{125}{7.5*10^3}[/tex]
[tex]I=\frac{1}{60}A[/tex]