8. A galvanometer has a coil resistance of 100 ohms and gives a full scale deflection when the potential difference across it is 0.0002 V. What size resistor (with a value measured in ohms) is required to make a 1.0 A. full scale deflection ammeter?
To solve this problem, we need to calculate the value of the resistor required to convert the galvanometer into a full-scale deflection ammeter.
First, we'll analyze the given information:
- The coil resistance of the galvanometer is 100 ohms.
- The potential difference (voltage) across the galvanometer for a full-scale deflection is 0.0002 V.
Now, let's determine the resistance of the resistor needed to convert the galvanometer into a 1.0 A full-scale deflection ammeter.
We know that ammeters are connected in parallel to the circuit element through which the current is to be measured. In order to ensure that the ammeter has minimal impact on the circuit, it needs to have very low resistance compared to the resistance of the circuit element.
According to Ohm's Law, the current flowing through a resistor is given by the equation:
I = V / R
Where:
- I is the current flowing through the resistor,
- V is the voltage across the resistor,
- R is the resistance of the resistor.
In this case, we want the ammeter to have a full-scale deflection of 1.0 A. Therefore, the current (I) needs to be 1.0 A.
We can rearrange Ohm's Law to solve for the resistance (R):
R = V / I
Plugging in the values:
V = 0.0002 V (potential difference across the galvanometer)
I = 1.0 A (desired current for ammeter)
R = 0.0002 V / 1.0 A
R = 0.0002 ohms
However, the resistance value of 0.0002 ohms is too small for practical use. It's more realistic and convenient to convert this value to a larger unit such as ohms (Ω). To do this, we can multiply the value by a multiplier of 1,000 to convert milliohms (mΩ) to ohms (Ω).
R = 0.0002 ohms * 1,000
R = 0.2 ohms
Therefore, a resistor with a resistance of 0.2 ohms is required to make a 1.0 A full-scale deflection ammeter with the given galvanometer.
current for meter: .0002/100=2E-5 amps
so the shunt should carry .(1-2E-5) amps
and the resistance =.0002/(1-2E-5) ohms
2E-3/(.99999) ohms