# Graph of current against time

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

## Answers and Replies

Andrew Mason
Science Advisor
Homework Helper
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:

$$\dot Q + Q/RC = 0[/itex] AM What is the equation for current as a function of time? Hint: it is the solution to the differential equation: $V_C + V_R = 0$ where $V_R = IR = R(dQ/dt)$ and $V_C = Q/C$ So: [tex]\dot Q + Q/RC = 0[/itex] AM Is there any other way, because this is precalculus physics? Andrew Mason Science Advisor Homework Helper Is there any other way, because this is precalculus physics? The solution is: [tex]Q = Q_0e^{-\frac{t}{RC}}$$

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

Last edited:
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 cant 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
Science Advisor
Homework Helper
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