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Help needed in finding current
- Thread starter mguptamel
- Start date
Your simulation is for 5V and you stated 10V earlier.
Here is my solution at 10V. See the diagram attached.
(1) Write KVL around the loop for current I1.
10 - 200I1 - 200I2 - 100I1 = 0
300I1 + 200I2 = 10
(2) Write KVL around the loop for current I2.
10 - 200I1 - 200I2 - 0.7 - 100I2 = 0
200I1 + 300I2 = 9.3
(3) Now we have two equations in two unknowns which have to be solved simultaneously. You can do this many ways ... I was lazy and punched them into an on-line calculator. The results are,
I1 = 22.8mA
I2 = 15.8mA
Here is my solution at 10V. See the diagram attached.
(1) Write KVL around the loop for current I1.
10 - 200I1 - 200I2 - 100I1 = 0
300I1 + 200I2 = 10
(2) Write KVL around the loop for current I2.
10 - 200I1 - 200I2 - 0.7 - 100I2 = 0
200I1 + 300I2 = 9.3
(3) Now we have two equations in two unknowns which have to be solved simultaneously. You can do this many ways ... I was lazy and punched them into an on-line calculator. The results are,
I1 = 22.8mA
I2 = 15.8mA
Attachments
I cheated. I built the circuit and measured it.
That drawing sucked! Anyway, the voltage after the 200 ohm resistor was 2.29 volts and the voltage between the diode and the 100 ohm resistor was 1.593 volts. After that, E=IR for 15.93 ma on the right and 22.9 ma on the left.
That drawing sucked! Anyway, the voltage after the 200 ohm resistor was 2.29 volts and the voltage between the diode and the 100 ohm resistor was 1.593 volts. After that, E=IR for 15.93 ma on the right and 22.9 ma on the left.
Attachments
#12, its nice when the calculation, the simulation, and the REAL circuit all agree!
MGUPTAMEL,
You can use any circuit analysis method you want as long as you include the diode in the circuit as a voltage source and get its polarity right. In circuit analysis there is always more than one way to solve a problem. But usually one or the other method is easier. People also have their favorites that they feel comfortable with. As an exercise, you could try to solve this problem using nodal analysis. Write the current equatins using KCL and try to solve.
MGUPTAMEL,
You can use any circuit analysis method you want as long as you include the diode in the circuit as a voltage source and get its polarity right. In circuit analysis there is always more than one way to solve a problem. But usually one or the other method is easier. People also have their favorites that they feel comfortable with. As an exercise, you could try to solve this problem using nodal analysis. Write the current equatins using KCL and try to solve.
Thanks all for helping me out earlier.
In attached screenshot, I have 2 example circuit. Circuit on left is with Vout where as circuit on right is Vout attached to a diode.
In left circuit, Vout (top) is 8V. As soon as I connect Vout to a diode (in right circuit), Vout has changed. The question is Vout changes depending on how I am completing the circuit. Textbooks refer to Vout a lot. If Vout is changeable, why Vout is referenced in textbooks?
In attached screenshot, I have 2 example circuit. Circuit on left is with Vout where as circuit on right is Vout attached to a diode.
In left circuit, Vout (top) is 8V. As soon as I connect Vout to a diode (in right circuit), Vout has changed. The question is Vout changes depending on how I am completing the circuit. Textbooks refer to Vout a lot. If Vout is changeable, why Vout is referenced in textbooks?
Attachments
Something has to be used to describe the output voltage of a circuit, so it's called, "Vout".
The problem here is that the load on a circuit always affects the circuit because it becomes part of the circuit when you attach it. Under optimum conditions, the load is insignificant compared to the circuit it is attached to. Example: high input impedance opamps. In this case, you have attached a diode which characteristically passes all voltage greater than about .7 volts. Above .7504 volts (as per your post) the diode is essentially a short circuit. If you really, really, tried to get it to support 4 volts, it would melt.
The problem here is that the load on a circuit always affects the circuit because it becomes part of the circuit when you attach it. Under optimum conditions, the load is insignificant compared to the circuit it is attached to. Example: high input impedance opamps. In this case, you have attached a diode which characteristically passes all voltage greater than about .7 volts. Above .7504 volts (as per your post) the diode is essentially a short circuit. If you really, really, tried to get it to support 4 volts, it would melt.
The problem here is that the load on a circuit always affects the circuit
Total understandable. This is confusing me for quite a while because as soon as a load is added for low resistance, for example, voltage will drop. Now if output of 1 circuit is an input of 2nd circuit, 2nd to 3rd ... so on and so forth, wouldn't the last circuit be affected as calculations will be mostly depend on Vout unchanged starting from 1.
As you mentioned that diode is a short >= 0.7 volts. The example I presented was to get an idea. It's not real life circuit though.
One more question...my apologies for so many questions...as these questions are haunting me for quite a while.
In textbooks transistor circuits talk in term of voltages but not current. Why would that be? From voltage, for example, Vout in above example, one can never know what is output current unless ammeter is connected to it? Please correct me if I am wrong. If circuit relies on some sort of input current, Vout won't help (again correct me). An example would be transistor where beta = 100 x alpha. I am trying to understand AC circuit where Vout and Vin (AC signal) is used.
Total understandable. This is confusing me for quite a while because as soon as a load is added for low resistance, for example, voltage will drop. Now if output of 1 circuit is an input of 2nd circuit, 2nd to 3rd ... so on and so forth, wouldn't the last circuit be affected as calculations will be mostly depend on Vout unchanged starting from 1.
As you mentioned that diode is a short >= 0.7 volts. The example I presented was to get an idea. It's not real life circuit though
. One more question...my apologies for so many questions...as these questions are haunting me
for quite a while.In textbooks transistor circuits talk in term of voltages but not current. Why would that be? From voltage, for example, Vout in above example, one can never know what is output current unless ammeter is connected to it? Please correct me if I am wrong. If circuit relies on some sort of input current, Vout won't help (again correct me). An example would be transistor where beta = 100 x alpha. I am trying to understand AC circuit where Vout and Vin (AC signal) is used.
1) yes, the last stage will be affected by errors in all the stages before it.
What you do to fix this is to design the circuit as if the load was already attached. As I have said before, most circuits are designed backwards...from the output to the input. You start with what the load is going to be and design the circuit to drive that load. In the example above, let us assume that the load is a 1k resistor and you expect to see 8 volts on it when you are done. You started out thinking that 12V through (3) 100 ohm resistors will produce 4 volts and 8 volts for outputs.
What you have to do is draw the circuit again, but with the 1k resistor attached to the alleged 8 volt output. When you do the math (Ohm's Law) you find that the alleged 8 volts is really 7.5 volts. You assume that the output is right, so it must be 8 volts. Eight volts applied to a 1k resistor causes .008 amps to flow. Meanwhile, 8 volts applied to (2) 100 ohm resistors will cause .04 amps to flow. Add the load currents to get .048 amps. You need the top resistor to allow .048 amps when 4 volts are pushing electricity through it. From Ohm's Law, R=E/I so R = 4/.048
The top resistor must be 83.333 ohms to get the output you desire.
2) Textbooks talk about voltage, but not current, because voltage and current are interdependent. It's just a simplification to speak of volts instead of amps. It could be done the other way just as easily, but doing both at the same time would actually confuse you more. You can calculate the current by knowing the voltage across a resistor, and you are expected to do that.
Now I have come to the point where I am confused as to what more you need. I will quit now and you can ask another question if you want to.
What you do to fix this is to design the circuit as if the load was already attached. As I have said before, most circuits are designed backwards...from the output to the input. You start with what the load is going to be and design the circuit to drive that load. In the example above, let us assume that the load is a 1k resistor and you expect to see 8 volts on it when you are done. You started out thinking that 12V through (3) 100 ohm resistors will produce 4 volts and 8 volts for outputs.
What you have to do is draw the circuit again, but with the 1k resistor attached to the alleged 8 volt output. When you do the math (Ohm's Law) you find that the alleged 8 volts is really 7.5 volts. You assume that the output is right, so it must be 8 volts. Eight volts applied to a 1k resistor causes .008 amps to flow. Meanwhile, 8 volts applied to (2) 100 ohm resistors will cause .04 amps to flow. Add the load currents to get .048 amps. You need the top resistor to allow .048 amps when 4 volts are pushing electricity through it. From Ohm's Law, R=E/I so R = 4/.048
The top resistor must be 83.333 ohms to get the output you desire.
2) Textbooks talk about voltage, but not current, because voltage and current are interdependent. It's just a simplification to speak of volts instead of amps. It could be done the other way just as easily, but doing both at the same time would actually confuse you more. You can calculate the current by knowing the voltage across a resistor, and you are expected to do that.
Now I have come to the point where I am confused as to what more you need. I will quit now and you can ask another question if you want to.
