A milliammeter of resistance 2.5ohms and a full deflection of 50mA iS to be used to measure a potential difference of 50v. What's the resistance of the multipliers?
To determine the resistance of the multipliers, we need to consider the total resistance in the circuit.
First, let's start with the milliammeter. We know that the milliammeter has a resistance of 2.5 ohms and a full deflection of 50mA. This means that when the current is at its maximum value of 50mA, the voltage drop across the milliammeter will be:
V = IR
V = (50mA) * (2.5 ohms)
V = 0.05A * 2.5 ohms
V = 0.125V
Now, let's consider the potential difference to be measured, which is 50V. Since we don't want the milliammeter to bear the full voltage drop, we need to introduce multipliers in the circuit.
Let R be the resistance of the multipliers. The total resistance in the circuit will be the sum of the milliammeter resistance (2.5 ohms) and the resistance of the multipliers (R).
So, the total resistance in the circuit will be:
Total Resistance = Milliammeter Resistance + Multipliers Resistance
Total Resistance = 2.5 ohms + R
To determine the current flowing through the circuit, we can use Ohm's Law:
V = IR
Since the total potential difference across the circuit is 50V and the total resistance is 2.5 ohms + R, we have:
50V = I * (2.5 ohms + R)
Now, we know that at full deflection, the milliammeter has a current of 50mA. This current will also flow through the circuit, so:
50mA = I
Now, we can substitute the current value and solve for R:
50V = (50mA) * (2.5 ohms + R)
By rearranging the equation and solving for R, we get:
R = (50V - (50mA * 2.5 ohms)) / 50mA
R = (50V - 0.125V) / 0.05A
R = 49.875V / 0.05A
R ≈ 997.5 ohms
Therefore, the resistance of the multipliers should be approximately 997.5 ohms to measure a potential difference of 50V using the given milliammeter.
at the meter: 2.5ohms*.050a=.125volts.
for the series resistance then, voltage=50-.125
so the resistance put in series should be (50-.125)/.050 -2.5=995 ohms
check that