Explanation on Shunt resistor from meter

AI Thread Summary
Meter shunts are designed to divert a specific proportion of current away from the meter coil while allowing a corresponding amount to flow through the meter, ensuring accurate measurements. For example, if a meter requires 1 amp for full-scale reading, only half an amp will yield a half-scale reading, maintaining proportionality. The discussion highlights the importance of understanding current division in parallel resistors and applying Ohm's Law for clarity. Additionally, there are concerns about the tone of the conversation, emphasizing the need for respectful communication in technical discussions. Overall, the mechanics of shunt resistors and their role in accurate current measurement are central to the topic.
omerhassan
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I know meter shunts are used to direct most of the current away from the meter coil but my question is if you direct most of current away from the coil, how is the meter measuring the current flow accurately.

Im kinda slow in the head so no complex fried banana explanation. tnx
 
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Can someone explain this to be quick, or are you all also slow in the head
 
A meter shunt diverts an exact proportion of the total current away from the meter but also an exact proportion goes though the meter.

So, if it takes 1 amp to make the meter read full scale, it will take half an amp to make it read half scale. Similarly, the meter reading is always proportional to the actual total current.
 
omerhassan said:
Can someone explain this to be quick, or are you all also slow in the head

This comment is unacceptable. You can say what you like about your own intelligence, but it is insulting and outside Forum rules to assume anyone who doesn't choose to answer your question is stupid.

Answering questions is purely optional.
 
Vk6kro thank you, I don't mean stupid, I mean slow typers. But I still don't understand your explanation but thanks
 
Suppose the meter takes 1 mA and has a resistance of 900 ohms.
When 1 mA is flowing, the voltage across the meter is I * R or 0.001 amps * 900 ohms or 0.9 volts.

If you put a 100 ohm shunt across the meter, then the current through the shunt for a full scale meter reading would be 0.9 volts / 100 ohms or 9 mA.

So it now takes a total of 10 mA to get a full scale deflection on the meter.

Can you see that it takes 5 mA to get a half scale deflection on the meter?
 
Think of the meter as a resistor.

Surely you can solve how current divides between parallel resistors?Changing them to admittances helps visualize it.

Ohm's Law - it's more than just a suggestion.
 
omerhassan said:
Can someone explain this to be quick, or are you all also slow in the head

That can only be taken as an insult. "In the head" doesn't refer to typing speed.
Afiac, you can solve your own problems. Your main one is to learn to be polite when you need to know something and to apologise when you have offended someone.
 

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