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A 12.4-µF capacitor is connected through a 0.895-MΩ 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 0s=0, at 5.0s=2.70x10-4Cat10.0s=4.42x10-4Cat20.0s=6.21x10-4Candat100s=7.44x10-4C

(b) The charging current at the same instants are as follows: at localid="1668329306161" 0s=6.67x10-5A,at5.0s=4.30x10-5A,at10.0s=2.74x10-5Aat20s=1.11x10-5Aandat100s=8.25x10-9A

Step by step solution

01

What is given to us

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

q=εC1-e-tRC……………………….(i)

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

i=εRe-tRC

It is also given that the capacitance of C=12.4μFthe resistance connected through it is R=0.895×106Ωand the potential difference connected across the capacitor and the resistor is V =60.0V

02

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:ε=V

Let’s first calculate the time constant of the R-C circuit:

τ=RC=0.895×106Ω12.4×10-6F=11.1s

Now, putting the values ofε,C,τ into the equation (i) we get:

q(t)=(60.0V)12.4×1061et11.1sq(t)=7.44×104C1et11.1s

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

q(t=0)=7.44×104C(11)=0q(t=5.00)=7.44×104C1e511.1=22.70×104Cq(t=10)=7.44×104C1e1011.1=4.42×104Cq(t=20.0)=7.44×104C1e20111.=6.21×104Cq(t=100)=7.44×104C1e1011.1=7.44×104C

03

Calculating the current at the given instances

Putting the values of ε,T,τ, to get the instantaneous current in the circuit at anu time t:

i(t)=60.0V0.895×106Ωet11.1si(t)=6.74×105Aet11.1s

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

i(t=0)=6.74×105A(1)=6.74×105Ai(t=5)=6.74×105Ae511.1=4.30×105Ai(t=10)=6.74×105Ae1011.1=2.74×105Ai(t=20)=6.74×105Ae2011.1=1.11×105Ai(t=100)=6.74×105Ae10011.1=8.25×109A

04

The graphs of the given data

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

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