Forest RC circuits - Revision history

Foretony: /* Diff. Eq. sol. method #2 */

2011-01-19T03:51:43Z

Diff. Eq. sol. method #2

Oborn: /* Diff. Eq. sol. method #1 */

2008-04-03T18:28:09Z

Diff. Eq. sol. method #1

Oborn: /* Diff. Eq. sol. method #1 */

2008-04-03T18:26:47Z

Diff. Eq. sol. method #1

Oborn: /* Charging RC circuit */

2008-04-03T18:24:11Z

Charging RC circuit

Oborn: /* Diff. Eq. sol. method #1 */

2008-04-03T18:13:43Z

Diff. Eq. sol. method #1

Oborn: /* Diff. Eq. sol. method #2 */

2008-04-03T18:12:26Z

Diff. Eq. sol. method #2

Oborn: /* Diff. Eq. sol. method #2 */

2008-04-03T18:11:56Z

Diff. Eq. sol. method #2

Oborn: /* Diff. Eq. sol. method #2 */

2008-04-03T18:09:58Z

Diff. Eq. sol. method #2

Oborn: /* Charging RC circuit */

2008-04-03T18:04:45Z

Charging RC circuit

Oborn: /* Charging RC circuit */

2008-04-03T17:53:15Z

Charging RC circuit

@@ Line 107: / Line 107: @@
 :<math>\int_0^q dq = \int_o^t \frac{V_b}{R}e^{-t/RC} dt</math>
-: <math>q = \frac{V_b}{R} \left ( -RC e^{-t/RC}\right ) |_0^t</math>
+: <math>q = \frac{V_b}{R} \left . \left ( -RC e^{-t/RC}\right ) \right |_0^t</math>
 : <math>q = CV_b\left ( 1 - e^{-t/RC}\right ) = Q\left ( 1 - e^{-t/RC}\right ) </math>

← Older revision		Revision as of 18:28, 3 April 2008
Line 76:		Line 76:

	:<math>q=CV_b (1 - e^{-t/RC}) =Q (1 - e^{-t/RC})</math>		:<math>q=CV_b (1 - e^{-t/RC}) =Q (1 - e^{-t/RC})</math>
		+
		+	;Note
		+	:you can also show that <math>A=CV_b</math> if you apply the boundary condition that <math>I= \frac{dq}{dt} = 0</math> when <math>t= \infty</math>.

	== Diff. Eq. sol. method #2 ==		== Diff. Eq. sol. method #2 ==

@@ Line 70: / Line 70: @@
 : <math>\Rightarrow \frac{1}{RC} \left ( A  \right ) = \frac{V_b}{R}</math>
 : <math>A = C V_b</math>
 == Diff. Eq. sol. method #2 ==

@@ Line 66: / Line 66: @@
 You then substitute this trial solution into the differential equation and apply boundary conditions to determine the coefficients.
-: \frac{dq}{dt} =
+: <math>\frac{d}{dt} \left ( A + Be^{-t/RC} \right ) + \frac{1}{RC} \left ( A + Be^{-t/RC} \right ) = \frac{V_b}{R}</math>
 == Diff. Eq. sol. method #2 ==

@@ Line 62: / Line 62: @@
 A common and more general technique to solving such a differential equation is to use the solution from the discharging equation (the homogeneous diff. eq.)  as a trial solution by inserting constants.
-q_{trial} = A + Be^{-t/RC}
+:<math>q_{trial} = A + Be^{-t/RC}</math>
 You then substitute this trial solution into the differential equation and apply boundary conditions to determine the coefficients.
 == Diff. Eq. sol. method #2 ==

@@ Line 79: / Line 79: @@
 ;just copy the solution for a discharging capacitor and replace "q" with "I"
-: I= I_0e^{-t/RC}
+: <math>I= I_0e^{-t/RC}</math>
 where<math> I_0= \frac{V_b}{R}</math> is the current from the battery when the battery has just been turned on.
@@ Line 93: / Line 93: @@
 :<math>\int_0^q dq = \int_o^t \frac{V_b}{R}e^{-t/RC} dt</math>

← Older revision		Revision as of 18:09, 3 April 2008
Line 77:		Line 77:

	the differential equation for the charge(q) in a discharging capacitor is the same as the differential equation for the current I in a charging capacitor.		the differential equation for the charge(q) in a discharging capacitor is the same as the differential equation for the current I in a charging capacitor.
		+
		+	;just copy the solution for a discharging capacitor and replace "q" with "I"
		+	: I= I_0e^{-t/RC}
		+
		+	where<math> I_0= \frac{V_b}{R}</math> is the current from the battery when the battery has just been turned on.
		+
		+	;Note
		+	:as the capacitor charges up the current drawn from the battery will be less because the capacitor is working against the electromotive force of the battery.
		+
		+	the equation
		+
		+	:<math>I=\frac{V_b}{R}e^{-t/RC} = \frac{dq}{dt}</math> is just a differential equation for the charge
		+
		+	by directly integrating we have
		+
		+	:<math>\int_0^q dq = \int_o^t \frac{V_b}{R}e^{-t/RC} dt</math>

← Older revision		Revision as of 18:04, 3 April 2008
Line 59:		Line 59:
	the above differential equation is similar to the equation for a discharging capacitor except that instead of zero on the RHS we have the battery's voltage <math>V_b</math>. This is called a 1st order differential equation with a source term.		the above differential equation is similar to the equation for a discharging capacitor except that instead of zero on the RHS we have the battery's voltage <math>V_b</math>. This is called a 1st order differential equation with a source term.

−	A common technique to solving such a differential equation is to use the solution from the discharging equation (the homogeneous diff. eq.)	+	== Diff. Eq. sol. method #1 ==
		+	A common and more general technique to solving such a differential equation is to use the solution from the discharging equation (the homogeneous diff. eq.) as a trial solution by inserting constants.
		+
		+	q_{trial} = A + Be^{-t/RC}
		+
		+	You then substitute this trial solution into the differential equation and apply boundary conditions to determine the coefficients.
		+
		+	== Diff. Eq. sol. method #2 ==
		+
		+	An important feature of our differential equation for the charging capacitor is that the right hand side is a constant with time (perfect battery). If we differentiate with respect to time we get an equation that is analogous to the differential equation for a discharging capacitor.
		+
		+	: <math>\frac{d}{dt} \left (\frac{dq}{dt} + \frac{q}{RC} = \frac{V_b}{R} \right )</math>
		+	: <math>\frac{d^2q}{dt^2} + \frac{1}{RC}\frac{dq}{dt} = 0</math>
		+
		+	or
		+	: <math>\frac{dI}{dt} + \frac{I}{RC} = 0</math>
		+
		+	the differential equation for the charge(q) in a discharging capacitor is the same as the differential equation for the current I in a charging capacitor.

← Older revision		Revision as of 17:53, 3 April 2008
Line 58:		Line 58:

	the above differential equation is similar to the equation for a discharging capacitor except that instead of zero on the RHS we have the battery's voltage <math>V_b</math>. This is called a 1st order differential equation with a source term.		the above differential equation is similar to the equation for a discharging capacitor except that instead of zero on the RHS we have the battery's voltage <math>V_b</math>. This is called a 1st order differential equation with a source term.
		+
		+	A common technique to solving such a differential equation is to use the solution from the discharging equation (the homogeneous diff. eq.)