Faulty electricity homework question?

[z.js]

Hi,
Hoping someone will be able to help me out with a electricity homework question that sounds a bit unreasonable to me.

Homework Statement

Question 8 (Read the section after Q4 first!)

The I-V characteristics of a diode and a resistor to be used in a simple circuit with a varaible voltage power supply are shown in the graphs.

I have a linear graph for the resistor for which I = 200mA at V = 1 V
I have a graph for the diode for which I = 0 mA at V = 0, I = 20 mA at V = 1, I about 1000mA at V = 2 ... At V = 1.8, the graph is near-vertical.

The question:
The diode and resistor are placed in parallel and a variable voltage applied to them.
a. If a voltage of 4.0 V is applied to the combination, what current will flow through them both?

Homework Equations

Ohm's law, V=IR. Applicable always to Ohmic devices, and applicable to non-Ohmic devices only for constant V or I (R is always changing).

The Attempt at a Solution

If I extrapolate the graph of the I-V characteristic for the diode to V=4, the gradient approaches infinity. Thus the resistance of the diode approaches zero. (R = 1/gradient). Since V = IR, the voltage drop across the diode also approaches zero.

Therefore the effective resistance of the diode-resistor series pair is equal to the resistance of the resistor, 5 ohm.

At 4.0V, I = V/R = 4V/5Ohm = 0.8A = 800mA.

However, the answer states 400mA.

I have attached a photo of the page with the graphs.
If my working is wrong, I would really like to know where I went wrong!
Thanks!
Stephen

 

[Simon Bridge]

If I extrapolate the graph of the I-V characteristic for the diode to V=4, the gradient approaches infinity. Thus the resistance of the diode approaches zero.

Correct so far.

(R = 1/gradient). Since V = IR, the voltage drop across the diode also approaches zero.

Not correct.

Therefore the effective resistance of the diode-resistor series pair is equal to the resistance of the resistor, 5 ohm.

Not correct.

The resister and diode are in parallel.
You described a situation where the diode is conducting - therefore it can be replaced by a short circuit.

 

[dauto]

Is the connection parallel or series?

 

[Simon Bridge]

From post #1: explicit problem statement:

The question:
The diode and resistor are placed in parallel and a variable voltage applied to them.

... mind you, could be a typo.

 

[dauto]

From post #1: explicit problem statement:

... mind you, could be a typo.

I think it is a typo. that's why I asked.

 

[haruspex]

I think it is a typo. that's why I asked.

The question asks:

what current will flow through them both?

That implies it's in series. If in parallel, it should say through "the combination", or somesuch.

 

[z.js]

Sorry... it was a typo. The diode and resistor are in series, not parallel.
@Simon Bridge:
"(R = 1/gradient). Since V = IR, the voltage drop across the diode also approaches zero."
You said this was incorrect... could you elaborate on that?
"Therefore the effective resistance of the diode-resistor series pair is equal to the resistance of the resistor, 5 ohm."
Is this still wrong?

 

[haruspex]

Sorry... it was a typo. The diode and resistor are in series, not parallel.
@Simon Bridge:
"(R = 1/gradient). Since V = IR, the voltage drop across the diode also approaches zero."
You said this was incorrect... could you elaborate on that?
"Therefore the effective resistance of the diode-resistor series pair is equal to the resistance of the resistor, 5 ohm."
Is this still wrong?

The total drop across the pair is 4V, but you do not know how that is distributed.
Can you think of a way of superimposing the two graphs to represent this?

 

[z.js]

Well, in series, the current is constant across the circuit. So if I can find V for which both graphs have the same I, I guess that would be it... something to chew on... looks like I = 400mA for which both V are approximately equal to 2V!

Thanks guys!

 

[haruspex]

Well, in series, the current is constant across the circuit. So if I can find V for which both graphs have the same I, I guess that would be it... something to chew on... looks like I = 400mA for which both V are approximately equal to 2V!

Thanks guys!

Yes, that looks about right. But I ask again, can you think of a way of superimposing the graphs that would allow you to read the answer straight off?