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 (Ω). The voltage drop
across the capacitor is Q/C, where Q is
the charge (in coulombs, C), so in this case Kirchhoff's Law
gives
RI + Q/C =E(t)
But I = dQ/dt, so we have
R(dQ/dt) + 1/C (Q) = E(t)
Suppose the resistance is 10 Ω, the capacitance is 0.05 F, and...