Circuit Analysis: Current and Voltage Ranking

[orangefruit]

Homework Statement

Consider the circuit at right, consisting of two identical resistors (R1 and R2) and a red LED. The battery is ideal and maintains a constant voltage of +9.0 V (i.e., V1 - V2 = +9.0 V). An ideal voltmeter is connected across the LED and R2.

1. At point a, is the direction of the current to the right, to the left, or equal to zero?

2. Rank, from largest to smallest, the absolute values of the currents at points a, b, c, and d. If any of the currents are zero, state so explicitly.

a) a = b = c = d = 0.
b) a = b = c = d > 0.
c) a = b > c = d = 0.
d) a > b > c = d > 0.
e) a = c = d > b = 0.
f) b = c = d > a = 0.
g) c = d > b > a > 0.

3. Rank, from largest to smallest, the absolute values of the voltages across R1, R2, and the LED. If any of the voltages are zero, state so explicitly.

a) R1 = R2 = LED = 0.
b) R1 = R2 = LED > 0.
c) R1 = R2 > LED > 0.
d) R1 > R2 = LED = 0.
e) R1 > LED > R2 > 0.
f) LED > R1 = R2 = 0.

4. What, if anything, can be said about the absolute value of the reading on the voltmeter?

a) 0.0 V.
b) 9.0 V .
c) The absolute value of the voltmeter reading will be greater than 0.0 V and less than 9.0 V.

Homework Equations

V=IR. Loop Rules.

The Attempt at a Solution

For 1, I believe it is to the right due to battery placement.
2, e because no current will want to flow through the voltmeter.
3, d because the positioning of the LED is not allowing current flow.
4, 9, because it is parallel to the battery which is parallel to the components consisting of LED and resistor.

Any help will be appreciated. I hope that getting help will clear any misconceptions I have. Thank you!

 

[orangefruit]

Anyone? I feel like I could be wrong about the first and that will ruin all of my answers

 

[ehild]

Is the LED ideal? I mean it does not let current through when reverse biased? If it is not ideal, your answer to 1) is correct. If the LED is an ideal diode, the current at a is zero.

ehild

 

[orangefruit]

Yes, it is reverse-biased. It will not light.

 

[ehild]

Yes, but some very small current can flow through it if it is not an ideal diode. Was it mentioned during your class?

ehild

 

[orangefruit]

Ah, I see. This is introductory university class. I believe all is assumed to be ideal. And since nothing more has been mentioned, it should be ideal. Thank you for prompt responses!

 

[orangefruit]

I want to change my answers.
First: zero.
2. all equals zero
3. LED> R1=R2=0
4. 9

But i am now confused. These answers are strictly from observation in lab. however, we did not measure currents. Is zero current correct?

 

[ehild]

I want to change my answers.
First: zero.
2. all equals zero
3. LED> R1=R2=0
4. 9

But i am now confused. These answers are strictly from observation in lab. however, we did not measure currents. Is zero current correct?

If neither breakdown of the LED nor reverse current were mentioned your answers are correct.

But a LED as every diode let's some very law current flow when reverse biased. And the reverse current becomes high if the reverse voltage exceeds the break-down voltage of the LED. That break-down voltage is between 3.5 and 5 volts, so the LED certainly breaks down when reverse biased by 9 V. At the same time, the reverse current causes voltage drop across the resistors, so the voltage across the diode is 9-2R I(reverse).

You said about observations in lab. Have you really reverse biased a LED?


ehild

 

[orangefruit]

Thank you. Thank you for clearing up my lacking explanation that may cause misconceptions in the future. We constructed a 2.5-5 V circuit consisting of a bulb and LED in series. The bulb accounted for our current "measurement". We did a forward and reverse biased setup. In reverse, we saw that the voltage drop across the bulb was always zero and the LED was always equal to the powersource. Neither one lit. Only lit forward-biased when the LED hit its knee-voltage. So indeed, we did not break the LED. Thank you so much, you are wonderful!

 

[ehild]

Very nice that you did not connect 9V to the LED in reverse. It would have been an expensive lab if every student broke a LED

A teacher has to be careful with the data in a problem he gives to the students.

ehild

 

[orangefruit]

Hehe, yes :D