a galvanometer has a resistance 60 ohm and is of 5mA full scale deflection.how would you modify it so as to get 0-10V voltmeter
at 10 V the current is 5 mA
so the total resistance is
... 10 / .005 = 2 kohm
so a (2000 - 60) ohm resistor is used in series
To modify a galvanometer to function as a voltmeter with a range of 0-10V, you would need to add a shunt resistor in parallel to the galvanometer.
First, you need to determine the current that corresponds to the desired full-scale deflection of 10V. This can be calculated using Ohm's Law, where V is the voltage and R is the resistance:
I = V / R = 10V / 60Ω = 0.167A
Next, you need to calculate the resistance required for the shunt resistor. The shunt resistor will bypass most of the current, allowing only a fraction of it to pass through the galvanometer. The remaining current will be proportional to the voltage being measured.
If the galvanometer has a full-scale deflection of 5mA (0.005A), you can calculate the current going through the shunt resistor:
Ishunt = Igalvanometer - Igauge = 0.167A - 0.005A = 0.162A
Assuming that the voltage drop across the shunt resistor should be negligible compared to the voltage being measured (10V), you can calculate the resistance required for the shunt resistor using Ohm's Law:
Rshunt = Vshunt / Ishunt = 10V / 0.162A ≈ 61.73Ω
Therefore, you would need a shunt resistor with a resistance of approximately 61.73Ω. Connect this shunt resistor in parallel to the galvanometer in your circuit, ensuring that it can handle the power dissipation resulting from the current flowing through it.
After adding the shunt resistor, the galvanometer will act as a voltmeter with a range of 0-10V. The reading on the galvanometer will be proportional to the voltage being measured.