1)a galvanometer has a resistance of 40 ohm and is of 3 mA full-scale deflection.how would you modify it to act as a 0-5 A ammeter?
2)the galvanometer described in question 1 is to be converted into a 0-5 V voltmeter.
a)when the voltmeter is connected to a 5 V supply ,how great a current will need to flow through it?
b)what must the resistance be between the terminals of the voltmeter for that to happen?
No idea what this question is about.
V = iR
All else befoggles
1. Vf = If*Rg = 0.003 * 40 = 0.120 Volts.
R = Vf/I = 0.120/(0.50-0.003) = 0.241 Ohm resistor in parallel.
2a. I = If + Ir = 0.5A.
2b. Rt = E/I = 5/0.5 = 10 Ohms. = Total resistance.
1) To modify the galvanometer to act as a 0-5 A ammeter, we need to add a shunt resistor in parallel with the galvanometer. The shunt resistor will divert most of the current away from the galvanometer when higher currents are measured.
Let's calculate the value of the shunt resistor required:
Given:
Resistance of the galvanometer (G) = 40 ohm
Full-scale deflection current (Ig) = 3 mA
Desired full-scale current (Ia) = 5 A
We can calculate the shunt resistance (Rs) using the formula:
Rs = (G * Ia) / (Ig - Ia)
Substituting the given values:
Rs = (40 * 5) / (0.003 - 5)
= 200 / (-4.997)
≈ -40.038 ohm
Since it is not practical to have a negative resistance, we take its absolute value:
Rs ≈ 40.038 ohm
So, to modify the galvanometer into a 0-5 A ammeter, we need to add a shunt resistor of approximately 40.038 ohms in parallel with the galvanometer.
2) a) When the voltmeter is connected to a 5 V supply, we need to determine the current flowing through it. To find that, we can use Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R).
Using this formula, we can rearrange it and find:
I = V / R
Given:
Voltage (V) = 5 V
Resistance of the galvanometer (G) = 40 ohm
Substituting the given values:
I = 5 / 40
= 0.125 A
Therefore, a current of 0.125 A will need to flow through the voltmeter.
b) To determine the resistance required between the terminals of the voltmeter, we can use Ohm's Law again. Rearranging the formula, we get:
R = V / I
Given:
Voltage (V) = 5 V
Current (I) = 0.125 A
Substituting the given values:
R = 5 / 0.125
= 40 ohm
So, the resistance between the terminals of the voltmeter needs to be 40 ohms for a current of 0.125 A to flow through it.