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Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro

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48 LESSON 6. EXPONENTIAL RELAXATION<br />

Figure 6.3: Schematic of RC circuit used <strong>in</strong> example 6.2.<br />

C<br />

a<br />

V<br />

+<br />

R<br />

b<br />

6. The total current through any node (a po<strong>in</strong>t where multiple wires<br />

come together) must sum to zero (Kirchoff’s Law).<br />

If we were to measure the voltage between po<strong>in</strong>ts a and b <strong>in</strong> an RC circuit<br />

as illustrated <strong>in</strong> fig 6.3, these rules tell us first that the capacitor should<br />

cause the voltage along the left branch to fluctuate accord<strong>in</strong>g to<br />

i C = C dV ab<br />

dt<br />

(6.25)<br />

where i C is the current through the capacitor. If i R is the current through<br />

the resistor, then the voltage drop through the lower side of the circuit is<br />

Solv<strong>in</strong>g for the current through the resistor,<br />

V ab = V batt + i R R (6.26)<br />

i R = V ab − V batt<br />

R<br />

S<strong>in</strong>ce the total current around the loop must be zero, i C + i R = 0,<br />

(6.27)<br />

Dropp<strong>in</strong>g the ab subscript,<br />

0 = i R + i C = V ab − V batt<br />

R<br />

dV<br />

dt = V batt − V<br />

RC<br />

+ C dV ab<br />

dt<br />

(6.28)<br />

(6.29)

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