Find the charge flowing through a device if the current

Sapewa

Sapewa

Answered question

2021-12-21

Find the charge flowing through a device if the current is:
a) i(t)=3A, q(0)=1C
b) i(t)=(2t+5)mA,
q(0)=0(c)i(t)=20
cos(10t+π6)μA, q(0)=2μC(d)A,i(t)=10e30tsin40tA, q(0)=0

Answer & Explanation

scoollato7o

scoollato7o

Beginner2021-12-22Added 26 answers

Let's first define that:
q(t)=q(0)+t0i(t)dt
Solving for a
q(t)=t03dt+1=3tt0+1=(3t+q)c
Solving for b
i(t)=(2t+5)mA, q(0)=0
q(t)=t0(2t+5)dt+0=
(2t22+5t)Bigt0=(t2+5t)mC
Solving for c
i(t)=20cos(10t+π6)μA, q(0)=2μc
q(t)=t020cos(10t+π6)dt+2=
2010sin(10t+π6)t0+2=
2sin(10t+π6)t0+2=
2sin(10t+π6)2sin(10(0)+π6)+2=
2sin(10t+π6)12+2=
(2sin(10t+π6))+1)μC
Solving for d
i(t)=(10e30tsin40t)A, q(0)=0
q(t)=100te30tsin40tdt+0
Solving this integration by parts we have

MoxboasteBots5h

MoxboasteBots5h

Beginner2021-12-23Added 35 answers

Step 1
a) q(t)=i(t)dt+q(0)=(3t+1)C
b) q(t)=(2t+s)dt+q(v)=(t2+5t)mC
c) q(t)=20cos(10t+π6)+q(0)=(2sin(10t+π6)+1)μC
d) q(t)=10r30tsin40t+q(0)=10e30t900+1600(30sin40t40cost)
=e30t(0.16cos40t+0.12sin40t)C
alenahelenash

alenahelenash

Expert2023-05-14Added 556 answers

a) q(t)=q(0)+0ti(τ)dτ=1C+0t3Adτ=1C+3A·t
b) q(t)=q(0)+0ti(τ)dτ=0+0t(2τ+5)mAdτ=0+[τ2+5τ]0t=t2+5t
c) q(t)=q(0)+0ti(τ)dτ=2μC+0t20cos(10τ+π6)μAdτ
Using the integral identity cos(ax+b)dx=1asin(ax+b)+C, we can evaluate the integral:
q(t)=2μC+[2010sin(10τ+π6)]0t=2μC+2sin(10t+π6)
d) q(t)=q(0)+0ti(τ)dτ=0+0t10e30τsin(40τ)Adτ
Using integration by parts, with u=sin(40τ) and dv=10e30τdτ, we have du=40cos(40τ)dτ and v=130e30τ. Applying the integration by parts formula:
q(t)=130e30τsin(40τ)|0t0t130e30τ·40cos(40τ)dτ
q(t)=130e30tsin(40t)+430te30τcos(40τ)dτ
Using the integral identity eaxcos(bx)dx=eaxa2+b2(acos(bx)+bsin(bx))+C, we can evaluate the remaining integral:
q(t)=130e30tsin(40t)+43·e30τ(30)2+402[30cos(40τ)+40sin(40τ)]0t
q(t)=130e30tsin(40t)+43·e30t(30)2+402[30cos(40t)+40sin(40t)]0t
Simplifying further:
q(t)=130e30tsin(40t)+43·e30t900+1600[30cos(40t)+40sin(40t)]
q(t)=130e30tsin(40t)+43·e30t2500[30cos(40t)+40sin(40t)]
q(t)=130e30tsin(40t)+43·e30t2500·(30cos(40t)+40sin(40t))
xleb123

xleb123

Skilled2023-05-14Added 181 answers

Step 1:
a) Given the current as i(t)=3A and initial charge q(0)=1C, we can find the charge flowing through the device. The charge is given by the integral of the current with respect to time:
q(t)=i(t)dt
Substituting the given current, we have:
q(t)=3dt=3t+C
To find the constant of integration, we use the initial charge q(0)=1C. Substituting this value:
1=3(0)+CC=1
Therefore, the charge flowing through the device is:
q(t)=3t+1C
Step 2:
b) For the current i(t)=(2t+5)mA and initial charge q(0)=0C, the charge flowing through the device is given by:
q(t)=i(t)dt
Substituting the given current:
q(t)=(2t+5)dt=t2+5t+C
Using the initial charge q(0)=0C:
0=(0)2+5(0)+CC=0
Hence, the charge flowing through the device is:
q(t)=t2+5tC
Step 3:
c) With the current i(t)=20cos(10t+π6)μA and initial charge q(0)=2μC, the charge flowing through the device can be obtained as:
q(t)=i(t)dt
Substituting the given current:
q(t)=20cos(10t+π6)dt=2010sin(10t+π6)+C
Using the initial charge q(0)=2μC:
2=2010sin(π6)+CC=2103
Therefore, the charge flowing through the device is:
q(t)=2010sin(10t+π6)+2103μC
Step 4:
d) Lastly, for the current i(t)=10e30tsin40tA and initial charge q(0)=0C, the charge flowing through the device is given by:
q(t)=i(t)dt
Substituting the given current:
q(t)=10e30tsin40tdt=11300e30t(40cos40t30sin40t)+C
Using the initial charge q(0)=0C:
0=11300(40cos030sin0)+CC=0
Therefore, the charge flowing through the device is:
q(t)=11300e30t(40cos40t30sin40t)C

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