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If An ammeter shunt has a voltage drop of 50 mega volts when 50 Amps of current flows through it what is the resistance of the shunt?
R = E / I
= (50 x 106) / (50)
= 1 megohm.
Strange for a "shunt". Must be across one heck of a meter movement !
It looks like the question was misworded. Instead of 50 megavolts, perhaps it should have been 50 millivolts. In that case the meter/shunt impedance would have been 0.001 ohms. In any case, the actual value of the shunt resistor would depend on the impedance of the meter itself. In the latter (assumed) case, this is probably negligible, so the shunt does appear to be 0.001 ohms.
What else can I help you with?
Why must you make current flow through the meter in order to measure it?
You don't. ...unless you want to directly measure the current in a circuit branch. That's the purpose of an ammeter. You can also use a volt meter if you know the resistance of a resistor in that branch to determine current (assuming DC circuit here) - current = voltage / resistance. This may be more useful for circuitry that is on a breadboard, since inserting an ammeter may not be practical.
What is the relationship among voltage current and resistance in a circuit?
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)
What happens to voltage when resistance increases?
Ohm's law states that "The current is directly proportional to the applied EMF (voltage) and inversely proportional to the resistance in the circuit." <<>> if resistor exists, resistance decreases according to ohm's law, current is directly proportional to voltage and current is inversely proportional to resistance it means as current increases, voltage increases. resistance increases, current decreases so as voltage if there is no resistor, there should be no resistance except internal resistance of voltmeter and ammeter
If a toy car with a resistance of 2 ohms is connected to a 3v battery how much current flows through the car?
Ohm's Law: V = IR (voltage = current times resistance).Ohm's Law: V = IR (voltage = current times resistance).Ohm's Law: V = IR (voltage = current times resistance).Ohm's Law: V = IR (voltage = current times resistance).
What is internal resistance of battery?
The equivalent resistance you would have to place in series with an ideal battery (which of course does not exist) of the same voltage to get the same behavior (voltage drop with load) as the real battery has. It is a mathematical modeling technique to help in circuit analysis.
How does an ammeter work?
An ammeter is a low voltage voltmeter in parallel with a small resistance resistor. Current flow through the resistor creates a voltage drop across it which is then measured by the voltmeter.
What is an ideal ammeter?
An ideal ammeter is a device that measures electric current and has zero resistance, producing no voltage drop when connected in a circuit. This ensures that the current being measured is not affected by the presence of the ammeter itself, providing an accurate reading of the current flowing through the circuit.
Why are the readings on a ammeter always different?
The readings on an ammeter indicate the current being drawn by a load in a circuit. This load is basically a resistance to current flow. The higher the resistance, the lower the current. The supply voltage has a direct effect on current flow. The higher the voltage applied, the higher the current will be. So the readings will vary on the ammeter according to fluctuations in load and or resistance of the circuit and the applied voltage.
How do you figure out how much current is in an electrical wire?
If you know the voltage and resistance, then current = voltage divided by resistance. Otherwise, you can attach an ammeter into the circuit (in series).
How calculate unknown resistance?
-- Connect a source of known, small voltage across the ends of the unknown resistance. -- Measure the resulting current through the unknown resistance. -- Divide (small known voltage)/(measured current). The quotient is the formerly unknown resistance.
What does the size of a current depend upon?
One way to determine current is to measure it, with an ammeter. Another way is to calculate it using Ohm's law: current = voltage / resistance.
Why should you be the internal resistance of the very few Amitr?
ammeter connect in series in circuit to measure the current, if the Ammeter have a high resistance it would effect the voltage value because there will be a drop voltage over the internal resistance of the Ammeter, so we desgin the ammeters with very low resistance...
Why does an ammeter have a low resistance while a voltmeter has high resistance?
I am going to assume that you mean low "resistance" in an open circuit test and are performing this with a multimeter. An ammeter works by place a very small amount of resistance in series with a circuit and then measuring the Voltage drop across the resistance. The Voltage is directly proportional to the current as given in ohms law: E = I x R If you are measuring the resistance through the ammeter it will have a very low resistance and impedance.
What is the effect caused by ammeter resistance when an ammeter is inserted into a circuit to measure the current?
The effect the multimeter might have on the circuit when inserted to measure the current is to increase the circuit resistance and decrease the available voltage to the circuit. This is because the multimeter in amps or milliamps mode does have a small resistance which is not zero, so by Ohm's law, there is a voltage drop across the multimeter; small, but not zero. Usually this effect is small. One way to compensate is to start by measuring voltage, and then inserting a separate ammeter and adjusting the power supply to match the original voltage. Of course, the voltmeter must be downstream of the ammeter.
Why is it a very bad idea to connect an ammeter directly across a voltage source?
Connecting an ammeter directly across a voltage source can cause a short circuit, damaging the ammeter and potentially causing overheating or even a fire hazard. This is because an ammeter has a low resistance, which could cause a very high current to flow through the circuit, overwhelming the device and damaging it. It is important to always connect an ammeter in series with the circuit to measure current accurately and safely.
Why don't we use an ammeter in series?
You do use an ammeter in series. It has a very low resistance, which according to ohms law, makes it ideal for measuring current. You do not use a volt meter in series, as it has a very HIGH resistance and would not allow current to flow. You measure the voltage across a component (or components), and current through a circuit.
Why shouldn't ammeter be connected in parallel?
A: An ammeter actually is a voltmeter measuring the voltage drop across a very small shunt resistance. They can measure AC or DC, make sure the meter is rated for the anticipated current, and always connect in series.
