How Is the Initial Current Calculated in a Discharge Graph?

[Masafi]

Hi, I need help with the last part. I have put each part with the answers.

For b) ii) I want to know where does V=1.6 V come from?

(a) A 2200 µF capacitor is charged to a potential difference of 12 V and then discharged through an electric motor. The motor lifts a 50 g mass through a height of 24 cm.

(i) Show that the energy stored in the capacitor is approximately 0.16

E = ½ CV^2 = 0.5 × 2200 × 10(–6) F × (12 V)^2
E = 0.158 J

(ii) What is the efficiency of the electric motor in this situation?

E = mgh = 0.12 J
Efficiency = 0.12 ÷ 0.16 J = 0.75

(b) The capacitor is charged to 12 V again and then discharged through a 16  resistor.

(i) Show that the time constant for this discharge is approximately 35 ms.

t = RC = 35.2 s

---- I need help with ----


(ii) Sketch a graph of current against time for this discharge on the grid below. You should indicate the current at t = 0 and t = 35 ms.

Curve starting on I axis but not reaching t axis
I = 1.6 V / 16Ohms = 100 mA shown on axis
Curve passing through about 37 mA at t = 35 ms

 

[Andrew Mason]

What is the equation for current as a function of time? Hint: it is the solution to the differential equation:

$V_C + V_R = 0$ where $V_R = IR = R(dQ/dt)$ and $V_C = Q/C$

So:

[tex]\dot Q + Q/RC = 0[/itex]<br /> <br /> AM[/tex]

 

[Masafi]

What is the equation for current as a function of time? Hint: it is the solution to the differential equation:

$V_C + V_R = 0$ where $V_R = IR = R(dQ/dt)$ and $V_C = Q/C$

So:

[tex]\dot Q + Q/RC = 0[/itex]<br /> <br /> AM[/tex]

$$<br /> Is there any other way, because this is precalculus physics?$$

 

[Andrew Mason]

Is there any other way, because this is precalculus physics?

The solution is:

$$Q = Q_0e^{-\frac{t}{RC}}$$

so:

$$\dot Q = I = \frac{V}{R} = -\frac{Q_0}{RC}e^{-\frac{t}{RC}} = -\frac{V_0}{R}e^{-\frac{t}{RC}}$$

Can you graph that?

AM

 

[Masafi]

The solution is:

$$Q = Q_0e^{-\frac{t}{RC}}$$

so:

$$\dot Q = I = \frac{V}{R} = -\frac{Q_0}{RC}e^{-\frac{t}{RC}} = -\frac{V_0}{R}e^{-\frac{t}{RC}}$$

Can you graph that?

AM

Yeah, so at t=0, I = 12V/16ohms = 0.75A

But the answer uses 1.6V, which I am not sure where they get from...

"I = 1.6 V / 16 = 100 mA shown on axis "

Shouldn't it be 12V since it says:

"The capacitor is charged to 12 V again and then discharged through a 16ohms resistor"

The time constant is approximately 35s, as the capacitance is 2200 microF.

On the graph in the answer, they have Current = 37mA at t=35, and Current = 100mA at t=0?

Basically, I can't get how they get the orignal V to be 1.6V, rather than 12V - there was no mention of 1.6V in the question...

 

[Andrew Mason]

You are right. The answers given correspond to an initial voltage of 1.6 V, not 12. The correct answers are 750 mA and 277 mA.

AM