(2 points) A circuit contains an electromotive force (a battery), a capacitor wi
ID: 2895565 • Letter: #
Question
(2 points) A circuit contains an electromotive force (a battery), a capacitor with a capacitance of C farads (F), and a resistor with a resistance of R ohms (). The voltage drop across the capacitor is Q/C, where Q is the charge (in coulombs), so in this case Kirchhoff's Law gives dQ dt Since the current is I- *, we have dt C Suppose the resistance is 30 , the capacitance is 0.05F, a battery gives a constant voltage of E(t)-30V, and the initial charge is Q(0) - 0C Find the charge and the current at time t. Q(t) = I() =Explanation / Answer
R(dQ/dt) +(1/C)Q =E(t)
=>30(dQ/dt) +(1/0.05)Q =30 , Q(0)=0
=>30(dQ/dt) +20Q =30
=>(dQ/dt) +(2/3)Q =1
=>dQ +(2/3)Qdt =1dt
integrating factor =e(2/3)dt =e(2/3)t
multiply by e(2/3)t on both sides
=>dQe(2/3)t +(2/3)Qe(2/3)t dt =e(2/3)t dt
=>(Qe(2/3)t)'=e(2/3)t dt
=>(Qe(2/3)t)=e(2/3)t dt
=>(Qe(2/3)t)=(3/2)e(2/3)t +C
=>Q=(3/2) +Ce(-2/3)t
Q(0)=0
=>(3/2) +Ce(-2/3)*0=0
=>(3/2) +C=0
=>C=-(3/2)
=>Q=(3/2) -(3/2)e(-2/3)t
=>Q=(3/2)(1 -e(-2/3)t)
(dQ/dt) +(2/3)Q =1
=>(dQ/dt) =1-(2/3)Q
=>I =1-(2/3)(3/2)(1 -e(-2/3)t)
=>I =1-(1 -e(-2/3)t)
=>I =e(-2/3)t
so finally
Q(t)=(3/2)(1 -e(-2/3)t)
I(t) =e(-2/3)t
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