Question

An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has an internal resistance of 10.0 Ω and the open-circuit voltage across its terminals is 50.0 V. The leads have no appreciable resistance. At time t = 0, the switch is suddenly closed.
(a) What is the maximum current through the 25.0-Ω resistor?
(c) What is the maximum charge that the capacitor receives?

Answer 1

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

Answer 2

Answer:

(a) Maximum current through resistor is 1.43 A

(b) Maximum charge capacitor receives is $1.50\times 10^{-3}\text{ C}$.

Explanation:

(a)

In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, t, is given by

$I(t) = I_0e^{-t/\tau}$

Here, $I_0$ 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

$I = \dfrac{V}{R_\text{eff}}$

V is the emf of the battery and $R_\text{eff}$ is the effective resistance.

In this question, $R_\text{eff}$ = 10.0 Ω + 25.0 Ω = 35.0 Ω

$I = \dfrac{50.0\text{ V}}{35.0\ \Omega} = 1.43\text{ A}$

(b) The maximum charge is given

Q = CV

where C is the capacitance of the capacitor

$Q = (30.0\times10^{-6}\text{ F})(50.0\text{ V}) = 1.50\times 10^{-3}\text{ C}$