The figure shows a circuit containing an electromotive force, a capacitor with a capa where Q is the charge (in coulombs), so in this case Kirchhoff's Law gives RI + 2 - E(t). But I = dQ/dt (see this example from a previous section), so we have R. = E(t).

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The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of C farads (F), and a resistor with a resistance of R ohms (2). 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 +
= E(t).
But I = dQ/dt (see this example from a previous section), so we have
DP
R
dt
= E(t).
Suppose the resistance is 2 N, the capacitance is 0.01 F, the electromotive force is E(t) = 70 sin(60t) V, and the initial charge is Q = 0 C.
Find the charge Q at time t.
Q =
- e
10
Find the current I at time t.
-50t
I = 35e
Transcribed Image Text:The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of C farads (F), and a resistor with a resistance of R ohms (2). 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 + = E(t). But I = dQ/dt (see this example from a previous section), so we have DP R dt = E(t). Suppose the resistance is 2 N, the capacitance is 0.01 F, the electromotive force is E(t) = 70 sin(60t) V, and the initial charge is Q = 0 C. Find the charge Q at time t. Q = - e 10 Find the current I at time t. -50t I = 35e
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