Calculate Current for 5V Sinusoidal Input Voltage

AI Thread Summary
To calculate the current flowing through a coil with a 5V sinusoidal input and a 1.5V output, impedance must be known, as current cannot be determined from voltage alone. The discussion highlights that measuring only voltage is insufficient without knowing resistance or impedance, especially in AC circuits. The user has measured impedance at 740 ohms but is confused about varying L, C, and R values with frequency. It is emphasized that without a load on the secondary coil, there will be no current flow, and thus no power can be measured. The importance of proper measurement techniques and circuit design is also noted to ensure accurate results.
Vikas_Madhu
Messages
37
Reaction score
0
I am doing an experiment where am giving 5 volts sinusoidal input voltage to a coil and receiving 1.5 volts in other coil how can i calculate current flowing through it..? if i take current as the ratio of input voltage to impedance of circuit will it be correct...?
 
Physics news on Phys.org
Is that use consistent with what impedance is?
 
Simon Bridge said:
Is that use consistent with what impedance is?

I didnt understand what you are asking for
 
You say you are uncertain if it is OK to use impedance the way you describe to fine the current.
Well... what is the definition of impedance? How does the definition relate to voltage and current in a circuit?
Then you have your answer.
 
Vikas_Madhu said:
I am doing an experiment where am giving 5 volts sinusoidal input voltage to a coil and receiving 1.5 volts in other coil how can i calculate current flowing through it..?

Have you considered applying the law of conservation of energy?
 
CWatters said:
Have you considered applying the law of conservation of energy?

No...
 
With just the data given in the OP you cannot calculate the current.
Many different transformers can be built with the same ratio between input and output voltages and they will all have different currents through the coils.
 
What else information is required ..?
 
Vikas_Madhu said:
What else information is required ..?

It is not clear if you mean a theoretical calculation or you mean to measure things.
What is the purpose of your investigation?
The easiest way will be to measure the current itself.
 
  • #10
I am telling about practical case.. right now i do not have a device to measure it so, am asking what is the other way through which i can proceed to calculate it
 
  • #11
Well, you need to measure something. How would you measure the impedance?
Is there anything that you can measure, then? Measuring just the voltages is not enough.
At least from what you say. The description of your setup is very vague.
You don't even mention what current do you want to calculate. In which coil?
 
  • #12
I am operating at 7 MHz, 5v frequency sinusoidal voltage, which is given to primary coil from function generator. I am receiving 1.5 volts in the secondary coil i need to calculate the power efficiency for which i am thinking of calculating current... now you got my problem?
 
  • #13
The power efficiency of what? Of the transfer between the primary and secondary coil?
This will depend on the load in the secondary, right?
If the secondary is open circuit, there is no power transferred, is it?
But you may be looking at something else. Who knows?
 
  • #14
Vikas_Madhu said:
I am telling about practical case.. right now i do not have a device to measure it so, am asking what is the other way through which i can proceed to calculate it

I am operating at 7 MHz, 5v frequency sinusoidal voltage, which is given to primary coil from function generator. I am receiving 1.5 volts in the secondary coil i need to calculate the power efficiency for which i am thinking of calculating current... now you got my problem?

You are still not quite understanding that you cannot measure the current when only given the voltage


to measure current you need to know 2 other values Resistance and Voltage ( for a DC circuit) or in your case, an AC circuit, Impedance and Voltage.
You are still giving only one of those required values, the 5V AC.
You need to work out the impedance of the primary coil ( its AC resistance), then you can work out the AC current flowing in the primary coil. From there you will be able to move on to working out the current flowing in the secondary coil with your given Voltage (1.5V) across what ever resistance of load you have across the secondary coil.

regards
Dave
 
  • #15
Hi nasu

hopefully the "penny will drop" for him soon :)

D
 
  • #16
davenn said:
You are still not quite understanding that you cannot measure the current when only given the voltage


to measure current you need to know 2 other values Resistance and Voltage ( for a DC circuit) or in your case, an AC circuit, Impedance and Voltage.

Yeah you are correct that i gave only one value but my question was if we can calculate with impedance and voltage. I measured the values of L, C, R, Z with LCR meter and i found that all values are changing with frequency. at 7 Mhz i got the impedance as 740 ohms. so with this information i hope i can solve for current according to you. One more doubt i have is why L,C, R values are changing with frequency?
 
  • #17
One more doubt i have is why L,C, R values are changing with frequency?

because inductive and capacitive reactance change with frequency. That's the way it is :)

that is ... the value of the capacitor etc doesn't change ... its reactance changes for a given freq.

and as a result, your impedance will change with freq. So in a given circuit where you want a fixed input and output impedance, you need to calculate those reactances ( the resistor, capacitor and inductor) values for the freq of interest to maintain your required impedance


Dave
 
  • #18
nasu said:
The power efficiency of what? Of the transfer between the primary and secondary coil?

I am not sure of if am doing correct or not, i am checking the output from oscilloscope am not connecting it to any load. I just want to know how much power i am transmitting from primary coil and how much am receiving at the secondary coil. So can you tell if am going in proper way to get what am willing to get?
 
  • #19
Vikas_Madhu said:
I am not sure of if am doing correct or not, i am checking the output from oscilloscope am not connecting it to any load. I just want to know how much power i am transmitting from primary coil and how much am receiving at the secondary coil. So can you tell if am going in proper way to get what am willing to get?

As nasu and I have said, you are not going to have a current flowing in the secondary coil unless you have a load across it ... so no load = no current flow = no power to measure


Dave
 
  • #20
davenn said:
because inductive and capacitive reactance change with frequency. That's the way it is :)

that is ... the value of the capacitor etc doesn't change ... its reactance changes for a given freq.
Dave

Sir when i measured even i found change in L, C, R values of coil..
 
  • #21
Vikas_Madhu said:
Sir when i measured even i found change in L, C, R values of coil..

the values of the components cannot change ( other than a tiny bit of tolerance variation when/if they warm up) ... unless they are faulty

how were you measuring them ? what with?
signal or no signal supplied ?

how about drawing/supplying (a circuit diagram) of your layout ... a sharp and well lit photo wouldn't hurt either

Dave
 
  • #22
davenn said:
how were you measuring them ? what with?
signal or no signal supplied ?

how about drawing/supplying (a circuit diagram) of your layout ... a sharp and well lit photo wouldn't hurt either

Dave


I am measuring with LCR meter and also to my surprise i got C values as negative, i measured for four times even by changing the terminals also but the result is same

I have attached the measurement values with this mail.
 

Attachments

  • coil details.JPG
    coil details.JPG
    17.5 KB · Views: 390
  • #23
it will be just an error in the way you are measuring the component values

again I ask ...
how about drawing/supplying (a circuit diagram) of your layout ... a sharp and well lit photo wouldn't hurt either


Dave
 
  • #24
davenn said:
how about drawing/supplying (a circuit diagram) of your layout ... a sharp and well lit photo wouldn't hurt either


Dave

Sir am attaching the diagram of my experiment with this please look into it.
 
  • #25
Vikas_Madhu said:
Sir am attaching the diagram of my experiment with this please look into it.
one thing in diagram is i have not used impedance matching circuit and regulated power supply because i have problem in choosing the circuit correct for impedance matching
 
  • #26
davenn said:
Hi nasu

hopefully the "penny will drop" for him soon :)

D

Hello davenn. :smile:
 
  • #27
Vikas_Madhu said:
one thing in diagram is i have not used impedance matching circuit and regulated power supply because i have problem in choosing the circuit correct for impedance matching

I cannot see the diagram. Can you?
 
  • #28
I just want to know how much power i am transmitting from primary coil and how much am receiving at the secondary coil.

If the secondary circuit is open circuit then Isecondary = zero.

Power = voltage * current

so Psecondary is also zero.

Then I refer you to my to first reply.

Regarding the negative values for capacitance... Did you just apply an LCR meter set on the C range to the windings of the transformer? If so I suggest you read up on how an LCR meter works. Then think about what might happen if you measure an inductor with it set to the capacitance range.
 

Similar threads

Back
Top