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    1) State whether each of the diodes is coducting

    2)What is the reading on V2?

    3) Calculate the current flowing through the 1000 ohm resistance

    4)Calculate current flowing through D1
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    (Original post by xDaniel)
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    1) State whether each of the diodes is coducting

    2)What is the reading on V2?

    3) Calculate the current flowing through the 1000 ohm resistance

    4)Calculate current flowing through D1
    Forum guidelines ask that you don't just post a homework question with no other comment. You need to tell us how far you've got with this yourself. What it is you don't understand, and what exactly you need help with.
    We are not here to do your homework for you.
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    I need help with all the questions. Since current goes from positive to negative does it get blocked by D2 since the flat part is facing it and it goes through D1?
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    (Original post by xDaniel)
    Since current goes from positive to negative does it get blocked by D2 since the flat part is facing it and it goes through D1?
    Correct.

    The convention for current is a flow from +ve to -ve. NB for historical reasons only, this is opposite to the actual flow of electrons. (Confusing at first, but vitally important when trying to understand what is happening at the chemical and atomic levels).

    The flat part of the diode symbol is the cathode. The other end is the anode. You need to remember these names. For the diode to conduct the anode must be at a higher +ve potential than the cathode by around 0.6V to 0.7V or more for a silicon semiconductor device.

    The diode symbol is an arrow. Current flows in the direction of the arrow (when the above condition is met) but not the other which is why there is a bar across the cathode.


    2) Answer using Kirchoff's voltage law (KVL): the potential differences around a circuit sum to the supply voltage.

    3) Ohms law.

    4) Use Kirchoff's current law (KCL): the current entering a junction must equal the sum of the currents leaving that junction (and vice versa).
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    2) 5.35 V
    3) 5.35x10^-3 A
    4)I don't get how to find the total current entering that junction, it can't be 6/1200 because that's 5x10^-3 and I got 5.35 before..
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    (Original post by xDaniel)
    2) 5.35 V
    3) 5.35x10^-3 A
    Both correct.

    (Original post by xDaniel)
    4)I don't get how to find the total current entering that junction, it can't be 6/1200 because that's 5x10^-3 and I got 5.35 before..
    D2 is reverse biased so does not conduct. The voltmeters are very high resistance so virtually no current flows through them and can be assumed zero.

    That means all of the current flows through the 1K ohms resistor. i.e. 5.35mA.

    From KCL, the sum of the currents entering the 1K ohms resistor must also be 5.35x10-3A.

    So turn your attention to the D1 and the 200 ohms resistor in parallel.

    What is the current flowing through that resistor? You already have all the information you need for this.
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    So...
    Current through the resistor is 0.65/200= 3.25x10-3
    Then I just take 3.25 away from 5.35? (2.1x10^-3)

    I still don't get why 5.35 can be taken as the whole current entering the junction
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    (Original post by xDaniel)
    So...
    Current through the resistor is 0.65/200= 3.25x10-3
    Then I just take 3.25 away from 5.35? (2.1x10^-3)
    Correct.

    (Original post by xDaniel)
    I still don't get why 5.35 can be taken as the whole current entering the junction
    Read my previous answer carefully.

    Where else can the current flow?
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    Ah yeah the only thing it can go through is the resistor.

    Dumb question but if V=IR and I is 0 and R is infinity how does the voltmeter get a reading?
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    (Original post by xDaniel)
    Ah yeah the only thing it can go through is the resistor.

    Dumb question but if V=IR and I is 0 and R is infinity how does the voltmeter get a reading?
    Not a dumb question. A perfect voltmeter would not draw any current but since none are perfect, they will all pass a current - albeit a small current.

    It's all down to the design of the voltmeter and the accuracy of the measurement needed.

    In reality, every voltmeter will draw a finite amount of current most often in the uA range giving an equivalent input resistance of 10x106 ohms or so. Laboratory precision voltmeters can have a resistance of 10x109 ohms or more. (Old moving coil meters could be as a low as a few thousand ohms).

    So for all normal range measurements, the current through modern digital type voltmeters can be ignored.

    A digital votmeter works on the principle of threshold comparison:

    The input voltage is compared to an internally generated voltage ramping up from zero and with a constant linear slope. A counter is started running at the same time as the ramp voltage starts from zero. When the ramp voltage crosses the threshold set by the voltage being measured, the counter is stopped.

    The counter is calibrated such that a count of 1.000000 corresponds to a 1V threshold etc.

    In this way, the input resistance of the voltmeter can be made very high because it is not directly used for say a moving coil meter which needs a current to deflect the pointer.
 
 
 
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