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Can you make ct by putting low value shunt resistor across pt?
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Why wouldn't the shunt resistor in the ammeter protect the ammeter when the ammeter is connected in parallel?
Connecting an ammeter in parallel subjects that ammeter to the full supply voltage. The shunt resistor is not designed to sustain that value of voltage and will burn out. Also, the clue is in the word 'shunt' (which means 'in parallel') which means that the coil will also burn out!
Why thermocouple voltmeter is connected across shunt resistance and electronic voltmeter across capacitor?
There is no such a thing as a thermocouple volt meter. A analogue or digital millivolt meter or volt meter is connected across a shunt or parallel with the shunt to measure the current through the resistor. Say the resistor value = 1 Ohm, then by using the Ohm law formula to calculate the current, say the voltage (voltage drop), read on the volt meter is 1.5 Volt that is R*V = A that is, 1Ω*1.5V = 1.5 Amp. Any type of DC volt meter, analogue or digital can be used to measure the voltage across a capacitor if the value of the capacitor is large enough that reading will be true RMS. as long as the supply current (EMF Power) are larger than the load current.
What happens to the current when the ammeter position is changed in a series circuit?
it gets a bigger chance of giving wrong result. ANSWER Absolutely not in series the meter will read the same no matter where it located in the series circuit. the same meter cannot be placed in parallel to measure.
The value of the resistor has 5 volts across it and has 20mA flowing through it?
The value of a resistor with 5 volts across it and 20 milliamperes of current flowing through it is 250 ohms. Ohm's Law: E = I R R = E/I
Explain the Potential across the resistor during the charging and discharging process of the capacitor?
The voltage across a capacitors given as a time constant t= 63% the resistor value multiply buy capacitor value. it doesn't matter if it goes more or less negative it will follow this function
How do you calculated the current shunt?
Write down the Ohm's law expression of "V = I * R" where "V" is the voltage drop across shunt resistor, "I" is the current flowing through shunt and "R" is the shunt resistance.Substitute value of voltage "V" and current "I" in the Ohm's law expression. For example, if voltage across shunt is 10 volts and current flowing through it is 1 ampere, then the expression is 10 = 100 * R.Divide the expression throughout by 100 to calculate the "R" value. Using a calculator, find the value of "R." From the example, the value of "R" will be 0.1 ohm, which is the value of shunt resistor.
Why wouldn't the shunt resistor in the ammeter protect the ammeter when the ammeter is connected in parallel?
Connecting an ammeter in parallel subjects that ammeter to the full supply voltage. The shunt resistor is not designed to sustain that value of voltage and will burn out. Also, the clue is in the word 'shunt' (which means 'in parallel') which means that the coil will also burn out!
Why the windings of an ammeter should be of thick wire?
because lots of current is likely to be flowing through them.however a shunt ammeter is easier to build than the direct ammeter as described above. in a shunt ammeter you use an ordinary meter wound with fine wire and put a small value shunt resistor across it to take almost all the current.
Why thermocouple voltmeter is connected across shunt resistance and electronic voltmeter across capacitor?
There is no such a thing as a thermocouple volt meter. A analogue or digital millivolt meter or volt meter is connected across a shunt or parallel with the shunt to measure the current through the resistor. Say the resistor value = 1 Ohm, then by using the Ohm law formula to calculate the current, say the voltage (voltage drop), read on the volt meter is 1.5 Volt that is R*V = A that is, 1Ω*1.5V = 1.5 Amp. Any type of DC volt meter, analogue or digital can be used to measure the voltage across a capacitor if the value of the capacitor is large enough that reading will be true RMS. as long as the supply current (EMF Power) are larger than the load current.
What happens to the current when the ammeter position is changed in a series circuit?
it gets a bigger chance of giving wrong result. ANSWER Absolutely not in series the meter will read the same no matter where it located in the series circuit. the same meter cannot be placed in parallel to measure.
The value of the resistor has 5 volts across it and has 20mA flowing through it?
The value of a resistor with 5 volts across it and 20 milliamperes of current flowing through it is 250 ohms. Ohm's Law: E = I R R = E/I
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 work resistor?
when a resistor is connected in a circuit it drop some voltage across it.when a circuit have large input voltage then by using a resistor of suitable value we get the desired voltage.
If an ammeter reads up to 1 ampere and its internal resistence is 0.81 ohms what value of shunt resistance is required to increase the range to 10 A?
Pull up a chair. An ammeter with an internal resistance of 0.81 ohms has a max current of 1 amp. We want to increase its range to 10 amps with a shunt resistance. Now focus on this. The shunt will be connected in parallel with the meter. (It's a shunt resistor, so that's what that means.) The max the meter can carry is 1 amp, so the shunt, which is in parallel with the meter, will have to carry 9 amps around the meter. That means the 1 amp through the meter will be added to the 9 amps of shunt current through the shunt resistor to give us the 10 amps of total current that was asked to be measured. Make sense? Review time. The meter carries 1 amp (it's max current) and the shunt carries 9 amps around the meter. That means the shunt has less resistance than the meter so it can carry all that extra current around the meter. How much less is the resistance? That's what will lead us to the answer to the question. We have 0.81 ohms in parallel with some smaller resistance, Rs, the value of the shunt resistor. Focus again. The shunt must carry 9 times as much current as the meter, so the shunt resistor's value must be 1/9th as much as the meter's. Make sense? Let's recap. The meter, with an internal resistance of 0.81 ohms is going to be 9 times as resistive as the resistance of the shunt. The meter will carry 1/9th as much current as the shunt, so the shunt, which carries 9 times the current of the resistor, will be able to carry that much more current because it's only 1/9th as resistive. The shunt resistance, Rs, is 1/9th the value of the internal resistance of the meter, RIm, and that makes the math easy. Rs = RIm / 9 = 0.81 ohms / 9 = 0.09 ohms The shunt will have to have a resistance value of 0.09 ohms. Let's check our work. A max of 1 amp through the meter, whose resistance is 0.81 ohms works out to 0.81 volts dropped across that meter. Em = Im x Rm = 1 amp x 0.81 ohms = 0.81 volts (voltage dropped across the meter) Our shunt will have the same identical voltage drop (it must have!) and 9 amps of current through it, right? Yes. We have the both those bits of data. Let's do the math. Rs = Es / Is = 0.81 volts / 9 amps = 0.09 ohms (the shunt's resistance is 0.90 ohms) Our work checks. And if you were wondering if the second approach could have been used as the primary means of solving the problem, the answer is, "Yes, it can." Either method will solve the problem, and the answer can be checked with the other approach.
Why does a Led have a 470 resistor connected to the LED?
A: That resistor is there to limit the current to the LED it can be any value if the voltage is decreased or increased or no resistor if the voltage across the led is equal to the forward voltage drop.
What is the procedure to improve the voltage range of voltmeter?
A: by adding a big value resistor from the source while measuring across a low value
How does understanding place value help you to subtract across zeros?
Understanding place value helps me to subtract across zeros by putting a \ on it
