Circuit Lab B/C

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Can someone explain this to me? The answer is 2 amps but I have no idea how to get that answer. https://drive.google.com/file/d/1FDlON7 ... sp=sharing

2 amps seems too high... that's a 12 ohm resistor, so that would require 24 volts across it. The only voltage sources are 12 V and 4 V, so I'm not sure how they got that answer either.

[mdv2o5]

Can someone explain this to me? The answer is 2 amps but I have no idea how to get that answer. https://drive.google.com/file/d/1FDlON7 ... sp=sharing

2 amps seems too high... that's a 12 ohm resistor, so that would require 24 volts across it. The only voltage sources are 12 V and 4 V, so I'm not sure how they got that answer either.

I agree; the answer is definitely not 2A. This problem is kind of tricky in that blindly following the KVL and KCL techniques will give you the wrong answer since you end up with current flowing backwards through one of the diodes.

After some more fiddling, I got 536mA flowing through the 12ohm resistor.

Also, as a side note, 0.75V is quite low for an LED forward voltage drop. For a red LED, a typical forward voltage is closer to 1.8V.

[mdv2o5]

I'm pretty new to this topic as well,
for the practical, do we just need to know how to use a multimeter or is there more.. will we have to calculate stuff from the made circuit?
also, are there any old practice tests for division b. I can only find ones from division C
thank you...

In the rulebook, the example task in 3.Part II.c.ii is a problem where you will have to calculate something from the circuit probably using a measurement that you need to make. The difficulty in this kind of problem probably doesn't come from actually using the multimeter but from knowing what values you need to do the calculation and then interpreting the circuit on the breadboard to make sure you're measuring the right thing. Of course, each exam will differ by how difficult the supervisor wants to make it.

[UTF-8 U+6211 U+662F]

Can someone explain this to me? The answer is 2 amps but I have no idea how to get that answer. https://drive.google.com/file/d/1FDlON7 ... sp=sharing

2 amps seems too high... that's a 12 ohm resistor, so that would require 24 volts across it. The only voltage sources are 12 V and 4 V, so I'm not sure how they got that answer either.

I agree; the answer is definitely not 2A. This problem is kind of tricky in that blindly following the KVL and KCL techniques will give you the wrong answer since you end up with current flowing backwards through one of the diodes.

After some more fiddling, I got 536mA flowing through the 12ohm resistor.

Also, as a side note, 0.75V is quite low for an LED forward voltage drop. For a red LED, a typical forward voltage is closer to 1.8V.

I also got 536 mA, so the answer key is most likely wrong.

[Crimesolver]

2 amps seems too high... that's a 12 ohm resistor, so that would require 24 volts across it. The only voltage sources are 12 V and 4 V, so I'm not sure how they got that answer either.

I agree; the answer is definitely not 2A. This problem is kind of tricky in that blindly following the KVL and KCL techniques will give you the wrong answer since you end up with current flowing backwards through one of the diodes.

After some more fiddling, I got 536mA flowing through the 12ohm resistor.

Also, as a side note, 0.75V is quite low for an LED forward voltage drop. For a red LED, a typical forward voltage is closer to 1.8V.

I also got 536 mA, so the answer key is most likely wrong.

Thanks everyone for the input! Would you mind explaining how you got that answer?

[knightmoves]

As drawn, the 12 and 6 ohm resistors are in parallel, so can be replaced by a single 4 ohm resistor.

The right hand loop on the picture is irrelevant, as the diode is reverse biased and does not conduct (proof left as an exercise for the reader).

So you have a simple loop circuit with a diode and 7 ohms of resistance, so i = 11.25/7 = 1.607A.

Now consider the 6 and 12 ohm resistors separately again. 1/3 of the current goes through the 12 ohm resistor (1/3 of the conductance), so answer = 0.536 A.

[UTF-8 U+6211 U+662F]

The right hand loop on the picture is irrelevant, as the diode is reverse biased and does not conduct (proof left as an exercise for the reader).

A little hint if you don't know where to start (Click to Show Hidden Text)

The easiest way I think you could go about seeing this is by marking the voltages for each node and seeing where you end up

[geniusjohn5]

My school doesn't have very many adequate supplies (my physics classroom only has crappy wires, switches, batteries, resistors, voltmeters, ammeters, and pigeon bulbs). What's the best way to prepare for the lab portion of the event?

[tangentline]

My school doesn't have very many adequate supplies (my physics classroom only has crappy wires, switches, batteries, resistors, voltmeters, ammeters, and pigeon bulbs). What's the best way to prepare for the lab portion of the event?

You can go to an electronics store (I used overpriced radioshack before) or shop online... Buy a breadboard, various resistance color coded resistors... An LED, decently sized capacitor... Just some stuff you can play around with. Perhaps buy a better multimeter. I don’t know if the lab gets too tricky... And some jumpers

[jaggie34]

As drawn, the 12 and 6 ohm resistors are in parallel, so can be replaced by a single 4 ohm resistor.

The right hand loop on the picture is irrelevant, as the diode is reverse biased and does not conduct (proof left as an exercise for the reader).

So you have a simple loop circuit with a diode and 7 ohms of resistance, so i = 11.25/7 = 1.607A.

Now consider the 6 and 12 ohm resistors separately again. 1/3 of the current goes through the 12 ohm resistor (1/3 of the conductance), so answer = 0.536 A.

Why would you ignore the right side, isn't the diode not in reverse bias in respect to the 4V source?