Suppose a circuit contains an electromotive force (a...

Suppose a circuit contains an electromotive force (a battery) that produces a voltage of E(t) volts (V), 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 RI + Q/C = E(t). Since I = dQ/dt, we have R dQ/dt + 1/C Q = E(t). Suppose the resistance is 30?, the capacitance is 0.05F, a battery gives a constant voltage of 30V, and the initial charge is Q(0) = 0 coulombs. Find the charge and the current at time t. Q(t) = I(t) =

Transcript

00:01 In the given question, the kirtcher's law is given as r -i plus q upon c is equal to e of t.

00:12 Since i is equal to d -q upon d -t, so we can write r into d -q upon d -t plus 1 -c to q is equal to e of t.

00:26 In the given question, r is equal to 30 -hom.

00:29 C is equal to 0 .05f and e of t is equal to 30 volt and q of zero is equal to zero we have to find q of t and i of t according to given information we can write 30 i plus q upon 0 .05 is equal to 30 on simplifying we will get 1 .5 i plus q is equal to 1 .5.

01:05 Now let yt denotes the change in circuit at time t, that is yt is equal to q of t.

01:22 This will implies y0 is equal to q0 which is equal to 0.

01:29 This is given in the question.

01:32 Since i is equal to dq all over d t, this will implies that i is equal to y -d -of -t -t.

01:42 Let this equation be star.

01:47 Now using these information's above equation will become 1 .5 y -dash t plus y of t is equals to 1 .5.

02:04 That is 1 .5 into d .y over dt plus y of t is equal to 1 .5.

02:16 Taking laplace on both the sides, 1 .5 laplace of y -d -t plus laplace of y of t plus laplace of y of t is equal to 1 .5 laplace of y -t is equals to 1 .5 laplace of.

02:35 1 this will be equals to 1 .5 as laplace of y of t minus y 0 plus la place of y of t is equal to 1 .5 since la place of 1 is equal to 1 upon s since la place of 1 is equal to 1 upon s and la place of y -d s into laplace of y of t is equal to s into laplace of y of t minus y of zero therefore 1 .5 into s into laplace of y of t minus zero since y zero is equal to zero plus laplace of y of t is equals to 1 .5 upon s.

03:49 This will be equal to 1 .5 s plus 1...

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