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2. Now we get rid of the differentials on both sides by integrating both sides. In algebra, you learned that you have to do the same thing to both sides of an equation so as not to change it. In physics, that means integrating over the same physical range on both sides. We make a choice here that at time t=0, we want there to be no charge on the plate, q(0) = 0. At some random time t in the future, the charge on the plate will simply be q(t). a. Integrate the dq side from q(0)-0 (just O is fine) to q(t). b. Integrate the dt side from t=0 to t. (Do you see what I mean by the same physical range on both sides?) c. The differentials dą and dt are now all added up and we have an equation we can solve for q(t). Solve it. d. Finally make the substitution Qmax = Ce. Rewrite the solution here.

2. Now we get rid of the differentials on both sides by integrating both sides. In algebra, you learned that you have to do the same thing to both sides of an equation so as not to change it. In physics, that means integrating over the same physical range on both sides. We make a choice here that at time t=0, we want there to be no charge on the plate, q(0) = 0. At some random time t in the future, the charge on the plate will simply be q(t). a. Integrate the dq side from q(0)-0 (just O is fine) to q(t). b. Integrate the dt side from t=0 to t. (Do you see what I mean by the same physical range on both sides?) c. The differentials dą and dt are now all added up and we have an equation we can solve for q(t). Solve it. d. Finally make the substitution Qmax = Ce. Rewrite the solution here.

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Now combining equation ( 1 ) and (2) we get e -()R - % = 0 e- % = R (#) = R (4) dq E – dt dq dt Ce-q dq C dt multiplying by dt on both sides and dividing by (q - Ce ) dt -dq CR q-Ce -dq q-Ce dt CR Now integrating both the sides, we get dq q-Ce - So CR •t dt Step 3 let q – Cɛ = z (let) dq = dz, now evaluating the limits when q = 0, z = -Cɛ, when q = q, z = q – Cɛ L # = - So CR -Cɛ dz dt Z. 19-Ce In z |-Ce CR In (q - Ce) — In (-Се) 3D CR q-Ce -Ce In CR Ce-q In Ce CR Cɛ-q e-1/CR Ce Ce – q = Ce e-1/CR q = Cɛ – Cɛ e-t/CR = Ce (1 – e-CR)
Transcribed Image Text:Now combining equation ( 1 ) and (2) we get e -()R - % = 0 e- % = R (#) = R (4) dq E – dt dq dt Ce-q dq C dt multiplying by dt on both sides and dividing by (q - Ce ) dt -dq CR q-Ce -dq q-Ce dt CR Now integrating both the sides, we get dq q-Ce - So CR •t dt Step 3 let q – Cɛ = z (let) dq = dz, now evaluating the limits when q = 0, z = -Cɛ, when q = q, z = q – Cɛ L # = - So CR -Cɛ dz dt Z. 19-Ce In z |-Ce CR In (q - Ce) — In (-Се) 3D CR q-Ce -Ce In CR Ce-q In Ce CR Cɛ-q e-1/CR Ce Ce – q = Ce e-1/CR q = Cɛ – Cɛ e-t/CR = Ce (1 – e-CR)
2. Now we get rid of the differentials on both sides by integrating both sides. In algebra, you learned that you have to do the same thing to both sides of an equation so as not to change it. In physics, that means integrating over the same physical range on both sides. We make a choice here that at time t=0, we want there to be no charge on the plate, q(0) = 0. At some random time t in the future, the charge on the plate will simply be q(t). a. Integrate the dq side from q(0)=0 (just 0 is fine) to q(t). Derivations We will begin by deriving the equations on the first page for charging and discharging capacitors in RC circuits. This requires solving a simple differential equation, but we will break it into simple steps in case you haven't had differential equations yet. It can be shown using Kirchhoff's rules (we will study them in a future lab) that for a charging capacitor in a simple RC circuit with a battery of e voltage, the rate of change of charge q(t) on the positive capacitor plate is given by: dq q dt R RC We don't write q(t) in the above equation because it becomes too cluttered and it is implied by the derivative anyway. e is the voltage AV of the battery. Our goal is to solve this equation for the function q(t). b. Integrate the dt side from t=0 to t. (Do you see what I mean by the same physical range on both sides?) Notice that the right-hand side becomes smaller as the charge on the plate q increases. This means it becomes increasingly harder to add charge to the plate; less charge is added as time increases. This equation is known as a first order, ordinary, linear, and separable differential equation for the function q(t). Quite a mouthful! First order means the equation only contains a first derivative, dq/dt. Ordinary means there are only full derivatives involved and one independent variable, instead of partial derivatives and several independent variables. Linear means there are no powers of our function or trigonometry functions involving our function, for example. Separable means it separates which is the first step to solving it! 1. Separate the differential equation by: a. Combining the two fractions on the right-hand side into one term (the common denominator is RC) с. The differentials dq and dt are now all added up and we have an equation we can solve for q(t). Solve it. b. Multiplying both sides by dt c. Dividing both sides q – Ce d. Finally make the substitution Qmax = Ce. Rewrite the solution here. тах d. You can tell that the differential equation has been separated because all the q's are on the dq side and all the t's (there weren't any) are on the dt side.
Transcribed Image Text:2. Now we get rid of the differentials on both sides by integrating both sides. In algebra, you learned that you have to do the same thing to both sides of an equation so as not to change it. In physics, that means integrating over the same physical range on both sides. We make a choice here that at time t=0, we want there to be no charge on the plate, q(0) = 0. At some random time t in the future, the charge on the plate will simply be q(t). a. Integrate the dq side from q(0)=0 (just 0 is fine) to q(t). Derivations We will begin by deriving the equations on the first page for charging and discharging capacitors in RC circuits. This requires solving a simple differential equation, but we will break it into simple steps in case you haven't had differential equations yet. It can be shown using Kirchhoff's rules (we will study them in a future lab) that for a charging capacitor in a simple RC circuit with a battery of e voltage, the rate of change of charge q(t) on the positive capacitor plate is given by: dq q dt R RC We don't write q(t) in the above equation because it becomes too cluttered and it is implied by the derivative anyway. e is the voltage AV of the battery. Our goal is to solve this equation for the function q(t). b. Integrate the dt side from t=0 to t. (Do you see what I mean by the same physical range on both sides?) Notice that the right-hand side becomes smaller as the charge on the plate q increases. This means it becomes increasingly harder to add charge to the plate; less charge is added as time increases. This equation is known as a first order, ordinary, linear, and separable differential equation for the function q(t). Quite a mouthful! First order means the equation only contains a first derivative, dq/dt. Ordinary means there are only full derivatives involved and one independent variable, instead of partial derivatives and several independent variables. Linear means there are no powers of our function or trigonometry functions involving our function, for example. Separable means it separates which is the first step to solving it! 1. Separate the differential equation by: a. Combining the two fractions on the right-hand side into one term (the common denominator is RC) с. The differentials dq and dt are now all added up and we have an equation we can solve for q(t). Solve it. b. Multiplying both sides by dt c. Dividing both sides q – Ce d. Finally make the substitution Qmax = Ce. Rewrite the solution here. тах d. You can tell that the differential equation has been separated because all the q's are on the dq side and all the t's (there weren't any) are on the dt side.
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