An 50mV meter has a resistance of 10ohms.a multiplier was inserted to producea range of 5V.?

Question

how can the multiplier be modified so that the new meter will have a range of 15V?

Answer 1

The 50 mV meter reads full scale when a current of 5 mA flows through its 10 ohm internal resistance.

I = E / R

0.050 V / 10 ohms = 0.005 amps.

The 5 volt multiplier is in series with the meter's 10 ohm internal resistance.  Thus, the multiplier resistance is:

(5 V - 0.050 V) / 0.005 amps = 990 ohms.

The 15 volt multiplier is then connected in series with the 990 ohm 5 volt multiplier.  Its resistance is, (15 V - 5 V - 0.050 V)  0.005 amps = 1,990 ohms

Answer 2

current sensitivity of meter is 50mV/10ohms = 5mA

for 5 volts, you need a resistor of 5v/5ma = 1000 ohms, less 10 or 990 ohms

for 15 volts you need 15v/5ma = 3000 ohms, less 10 or 2990 ohms.

To change from 5v to 15 v you can add the difference of 2000 ohms in series.

Answer 3

Divide your 50 mV by 10 ohms to get a full scale current of 5 mA.  Now divide 5V by 5 mA to get 1000 ohms (less 10) = 990 ohms.

15V / 5mA = 3000, - 1000 = 2000.  Add a 2 kiloohm resistor in series with the 990 ohm and the meter, to display 0-15V on a 0-50mV meter movement.

Answer 4

All of the math is good to know since you'll meet up with it on a test. A good reference manual for this sort of thing would be the ARRL radio amateurs handbook. A quick and dirty way to get your answer would be to get a couple of variable resistors. Connect them so that one main lead goes to the power source, and the slide contact goes to the meter. Put this in the positive lead of the meter, and the negative lead to the power supply.  Set the variable to full resistance, turn on the voltage source for the desired voltage, and slowly adjust the variable resistor (a potentiometer) until you arrive at a full scale reading. Turn off the power and measure the resistance across the 2 contacts you used. That will be the resistance you need to gain your series resistance for the desired voltage scale. Ohms law is a good law to use here, you would do very good to learn that tool because it is one of the best formulas that you'll ever use. For a first approximation, it is hard to beat, unless you really like to use more extensive math. Since I=E/R the 50mV divided by the 10 ohms of the meter resistance, that will give you the current needed to swing the meter to full scale. You need more resistance, so your formula would be the meter current found by my first formula, divided into the voltage that you need.