# Two quick questions - circuits

1. Jan 6, 2010

### mathman44

1. The problem statement, all variables and given/known data

#1 A power source produces a "sawtooth" wave of current, reversing direction at 1s. After 2s (The current starts at 0, reaches 1 amp at 1 second, reverses direction to -1 amp, and then reaches 0 again at 2 seconds), calculate the total energy supplied to a 5000 ohm resistor.

#2 Different measurements are taken by attaching various resistors R across a cell terminal and recording the voltages and corresponding currents I. From the graph, find

(a) the open circuit voltage of the cell.
(b) the short-circuit current
(c) the maximum power that the cell can deliver to an external resistance.

The graph is a negative linear line. V on the x-axis, I on the y-axis. The starting point is at () V, 3 amps) and the end point is at (1 V, 0 amps).

2. The attempt at a solution

#1 - Energy = C (charge) x V, so I'm tempted to say that energy is zero because the current is positive for the first second and negative for the second, thus they should cancel.

#2 - I'm not sure what the open/short circuits are referring to... could anyone clarify? As for c, power P = V*I, but at what point is this maximum?

Last edited: Jan 6, 2010
2. Jan 6, 2010

### rock.freak667

For the first one, use E=I2Rt to find the energy for the two directions.

3. Jan 6, 2010

### mathman44

The current is not constant, so how could I use that equation? Here are the graphs I described:

4. Jan 7, 2010

Bump.

5. Jan 7, 2010

### tiny-tim

Integrate: energy = ∫ power dt = ∫ I2R dt.

6. Jan 7, 2010

### RoyalCat

You know $$I(t)$$ over the relevant intervals. You said it yourself, the current is a function of time. You can't just say that $$\int I^2 dt = I^2 t$$
That's not true when $$I$$ is a function of time!

Use the differential form:
$$P=\frac{dE}{dt}$$ to rewrite: $$dE=I^2 R dt$$ and then integrate over the relevant intervals.

Last edited: Jan 8, 2010
7. Jan 7, 2010

### mathman44

So integrating both sides yields:

energy = ∫ I2R dt

What function am I integrating though? All I have is this graph :(...something to do with a Sin wave perhaps?

Last edited: Jan 7, 2010
8. Jan 7, 2010

### RoyalCat

Find the current as a function of time over the two relevant intervals. You drew a perfect diagram of a linearly rising current, now just derive its equation!

9. Jan 7, 2010

### mathman44

Can I approximate the sawtooth as two integrals,
I(t) = t for t=0 to t=1
and
I(t) = t - 2 for t=1 to t=2

Doing this and integrating, I get 0 for energy.

Last edited: Jan 7, 2010
10. Jan 7, 2010

### JaWiB

Square of the current!

11. Jan 7, 2010

### mathman44

Oops. After squaring I(t) for both integrals, I'm getting 10000/3 for E. Does this make sense? I was expecting zero.

12. Jan 7, 2010

### willem2

It's correct. The power supplied to a resistor is always positive if there is a nonzero current. if the power supplied could become negative, then the resistor would be supplying power to the battery. This power is equal to $I^2R$ wich is always positive as well.

13. Jan 7, 2010

### mathman44

Thanks, that makes sense.

Could someone confirm these answers for the second question?
a) 1 volt
b) 3 amp

for c)... The equation for I=3-3V so the power would be P=IV=3V-3V^2, right?

This has a maximum when dP/dV =0 = 3-6V or V=1/2.

so Pmax =(3-3(1/2))*(1/2)= 3/4?

14. Jan 8, 2010

### mmmboh

Haha I'm in your class! but yes I got the same answers as you did for that problem.