A ammeter is to be constructed using a galvanometer that has a 2.00 mA full-scale deflection and internal resistance of 50.0 ohms. It must be good enough to detect 40 mA, which is the max current. How would one do this? How would you make a voltmeter able to read the largest voltage drop(which is 14.0 v) in the circuit?
Please show work. Thank You
Va = Ia*Ra = 2*10^-3 * 50 = 0.10 Volts =
Voltage across ammeter @ full-scale.
Rp=Va/I = 0.1/(40-2)*10^-3=0.1/38*10^-3=
2.632 Ohms. = Resistor connected in
parallel with the ammeter to extend
current range to 40 mA.
Voltmeter:
Rs = (V-Va)/If = (14-0.10)/2*10^-3=6,950
Ohms = The resistor connected in series
with the ammeter.
Note: The 2.632-Ohm resistor is not used
with the voltmeter arrangement.
To construct an ammeter using a galvanometer, you need to connect a shunt resistor in parallel with the galvanometer to divert the excess current and limit the current flowing through the galvanometer.
To calculate the value of the shunt resistor, you'll need to use the formula:
Rs = (Ig - Ia) / Ia * Rg
Where:
Rs is the value of the shunt resistor
Ig is the full-scale deflection current of the galvanometer (2.00 mA)
Ia is the maximum current to be measured (40 mA)
Rg is the internal resistance of the galvanometer (50.0 ohms)
Plugging in the values:
Rs = (0.040 A - 0.002 A) / 0.002 A * 50.0 ohms
Rs = 0.038 A / 0.002 A * 50.0 ohms
Rs = 38 / 0.002 * 50.0 ohms
Rs = 38 * 50.0 ohms / 0.002
Rs = 1900 ohms
Therefore, you need to connect a 1900-ohm shunt resistor in parallel with the galvanometer to create an ammeter capable of measuring currents up to 40 mA.
Now, let's move on to constructing a voltmeter.
To construct a voltmeter, you need to connect a high-value resistor called a multiplier resistor in series with the galvanometer. The purpose of the multiplier resistor is to limit the current through the galvanometer and provide a suitable voltage range for measurement.
To determine the value of the multiplier resistor, you can use the following formula:
Rm = V / (Ig / K)
Where:
Rm is the value of the multiplier resistor
V is the maximum voltage to be measured (14.0 V)
Ig is the full-scale deflection current of the galvanometer (2.00 mA)
K is the sensitivity adjustment factor of the galvanometer (could be given or determined from its specifications)
Assuming the sensitivity adjustment factor (K) is 1, the calculation would be as follows:
Rm = 14.0 V / (0.002 A / 1)
Rm = 14.0 V / 0.002 A
Rm = 7000 ohms
Therefore, you need to connect a 7000-ohm multiplier resistor in series with the galvanometer to create a voltmeter capable of measuring voltage drops up to 14.0 V.
Remember to connect the positive terminal of the galvanometer to the positive side of the shunt resistor for the ammeter and the positive side of the multiplier resistor for the voltmeter. Also, make sure to observe correct polarity to get accurate measurements.