Answer :
Answer:
(a) Maximum current through resistor is 1.43 A
(b) Maximum charge capacitor receives is [tex]1.50\times 10^{-3}\text{ C}[/tex].
Explanation:
(a)
In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, t, is given by
[tex]I(t) = I_0e^{-t/\tau}[/tex]
Here, [tex]I_0[/tex] is the maximum current and τ represents time constant which is given by RC (the product of the resistance and capacitance).
The maximum current is given by
[tex]I = \dfrac{V}{R_\text{eff}}[/tex]
V is the emf of the battery and [tex]R_\text{eff}[/tex] is the effective resistance.
In this question, [tex]R_\text{eff}[/tex] = 10.0 Ω + 25.0 Ω = 35.0 Ω
[tex]I = \dfrac{50.0\text{ V}}{35.0\ \Omega} = 1.43\text{ A}[/tex]
(b) The maximum charge is given
Q = CV
where C is the capacitance of the capacitor
[tex]Q = (30.0\times10^{-6}\text{ F})(50.0\text{ V}) = 1.50\times 10^{-3}\text{ C}[/tex]
Answer:
Imax = 1.43A, Qmax = 1.5mC
Explanation:
Given the Capacitance of the capacitor, C = 30.0µF = 30×10-⁶F, E = 50V(emf)
The maximum current through the resistor is Imax = E/R = E/Re^(-t/RC) = E/R×e^(0) = E/R
E = 50V
R1 = 25Ω
r = R2 = 10Ω( resistance of the battery
R = R1 + R2 = 35Ω (series connection)
Imax = 50/35 =1.43A
Qmax = CE = 30×10-⁶×50 = 1.5×10-³C = 1.5mC