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)=10e−30tsin⁡40tA, q(0)=0

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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)=10e−30tsin⁡40tA, q(0)=0

scoollato7o · Accepted Answer

Let&#039;s first define that: q(t)=q(0)+∫t0i(t)dt Solving for a q(t)=∫t03dt+1=3t∣t0+1=(3t+q)c Solving for b i(t)=(2t+5)mA, q(0)=0 q(t)=∫t0(2t+5)dt+0= (2t22+5t)Big∣t0=(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)=(10e−30tsin⁡40t)A, q(0)=0 q(t)=10∫0te−30tsin⁡40tdt+0 Solving this integration by parts we have

MoxboasteBots5h · Accepted Answer

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)=∫10r−30tsin⁡40t+q(0)=10e−30t900+1600(−30sin⁡40t−40cos⁡t)  =−e−30t(0.16cos⁡40t+0.12sin⁡40t)C

alenahelenash · Accepted Answer

a) q(t)=q(0)+∫0ti(τ)dτ=1C+∫0t3Adτ=1C+3A·tb) q(t)=q(0)+∫0ti(τ)dτ=0+∫0t(2τ+5)mAdτ=0+[τ2+5τ]0t=t2+5tc) q(t)=q(0)+∫0ti(τ)dτ=2μC+∫0t20cos(10τ+π6)μAdτUsing 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+∫0t10e−30τsin(40τ)AdτUsing integration by parts, with u=sin(40τ) and dv=10e−30τdτ, we have du=40cos(40τ)dτ and v=−130e−30τ. Applying the integration by parts formula:q(t)=−130e−30τsin(40τ)|0t−∫0t−130e−30τ·40cos(40τ)dτq(t)=−130e−30tsin(40t)+43∫0te−30τcos(40τ)dτUsing the integral identity ∫eaxcos(bx)dx=eaxa2+b2(acos(bx)+bsin(bx))+C, we can evaluate the remaining integral:q(t)=−130e−30tsin(40t)+43·e−30τ(−30)2+402[−30cos(40τ)+40sin(40τ)]0tq(t)=−130e−30tsin(40t)+43·e−30t(−30)2+402[−30cos(40t)+40sin(40t)]0tSimplifying further:q(t)=−130e−30tsin(40t)+43·e−30t900+1600[−30cos(40t)+40sin(40t)]q(t)=−130e−30tsin(40t)+43·e−30t2500[−30cos(40t)+40sin(40t)]q(t)=−130e−30tsin(40t)+43·e−30t2500·(−30cos(40t)+40sin(40t))

xleb123 · Accepted Answer

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)dtSubstituting the given current, we have:q(t)=∫3dt=3t+CTo find the constant of integration, we use the initial charge q(0)=1C. Substituting this value:1=3(0)+C⟹C=1Therefore, the charge flowing through the device is:q(t)=3t+1CStep 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)dtSubstituting the given current:q(t)=∫(2t+5)dt=t2+5t+CUsing the initial charge q(0)=0C:0=(0)2+5(0)+C⟹C=0Hence, the charge flowing through the device is:q(t)=t2+5tCStep 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)dtSubstituting the given current:q(t)=∫20cos(10t+π6)dt=2010sin(10t+π6)+CUsing the initial charge q(0)=2μC:2=2010sin(π6)+C⟹C=2−103Therefore, the charge flowing through the device is:q(t)=2010sin(10t+π6)+2−103μCStep 4:d) Lastly, for the current i(t)=10e−30tsin40tA and initial charge q(0)=0C, the charge flowing through the device is given by:q(t)=∫i(t)dtSubstituting the given current:q(t)=∫10e−30tsin40tdt=11300e−30t(−40cos40t−30sin40t)+CUsing the initial charge q(0)=0C:0=11300(−40cos0−30sin0)+C⟹C=0Therefore, the charge flowing through the device is:q(t)=11300e−30t(−40cos40t−30sin40t)C

Find the charge flowing through a device if the current

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