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A 12.4-碌F capacitor is connected through a 0.895-M鈩 resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for t between 0 and 20 s

Short Answer

Expert verified

(a) The charge on the capacitor at the following times after the connections are made is

At 0 S = 0 C, at 5.0 S = 2.7010-4C, at10.0S=4.4210-4C at 20.0S=6.2110-4C, and at100S=7.4410-4C

(b) The charging current at the same instants are as follows:

at 0S=6.6710-5, at5.0S=4.3010-5A at10.0S=2.7410-5A , at 20S=1.1110-5A, and at100S=8.2510-9A

(c) The graphs are drawn below.

Step by step solution

01

Concept

A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. The effect of a capacitor is known as capacitance.

02

What is given to us

Charge on a charging capacitor as a function of time is given by:

q=蔚颁(1-e-tRC)..............................(1)

And the function between time and current in a circuit is given by:

i=Re-tRC

It is also given that the capacitance of C=12.4渭贵the resistance connected through it isR=0.895106 and the potential difference connected across the capacitor and the resistor is V = 60.0 V

03

Calculating the charge for the given instance

The constant potential difference connected to the capacitor and the resistance be our emf of the R-C circuit:=V

Let鈥檚 first calculate the time constant of the circuit:

=RC=(0.895106)(12.410-6F)=11.1S

Now, putting the values of into the equation (1) we get:

q(t)=(60.0V)(12.410-6F)1-e-t11.1Sq(t)=(7.4410-4C)1-e-t11.1S

Now, substituting the values of given times t to get the values of charges at each instant:

q(t=0)=(7.4410-4C)(1-1)=0q(t=5.00)=(7.4410-4C)1-e-5S11.1S=22.7010-4Cq(t=10)=(7.4410-4C)1-e-10S11.1S=4.4210-4Cq(t=20.0)=(7.4410-4C)1-e-20S11.1S=6.2110-4Cq(t=100)=(7.4410-4C)1-e-10S11.1S=7.4410-4C

04

Calculating the current at the given instances

Substituting the values of ,T and to get the instantaneous current in the circuit at any time t:

i(t)=60.0V0.895106e-t11.1Si(t)=(6.7410-5A)e-t11.1S

Now, we enter each value of the given time t and get the current at each of the instants:

i(t=0)=(6.7410-5A)(1)=6.7410-5Ai(t=5)=(6.7410-5A)e-5S11.1S=4.3010-5Ai(t=10)=(6.7410-5A)e-10S11.1S=2.7410-5Ai(t=20)=(6.7410-5A)e-20S11.1S=1.1110-5Ai(t=100)=(6.7410-5A)e-100S11.1S=8.2510-9A

05

The graphs are traced below

The graphs of the resultant part of the above instances for between and is given below:

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