A galvanometer of resistance 100 gives full scale deflection at 10 mA current. What should be the value of shunt so that it can measure a current of 100 mA?
To calculate the value of the shunt resistor, we can use the formula:
\(I_{\text{total}} = I_{\text{galvanometer}} + I_{\text{shunt}}\)
Given that the galvanometer has a resistance of 100 Ω and gives a full-scale deflection at 10 mA, we can calculate the shunt resistance value as follows:
\(I_{\text{galvanometer}} = 10 \, \text{mA}\)
\(R_{\text{galvanometer}} = 100 \, \Omega\)
\(I_{\text{total}} = 100 \, \text{mA}\)
Therefore, substituting the given values into the formula, we have:
\(100 \, \text{mA} = 10 \, \text{mA} + I_{\text{shunt}}\)
We can rearrange the equation and solve for the shunt current:
\(I_{\text{shunt}} = 100 \, \text{mA} - 10 \, \text{mA}\)
\(I_{\text{shunt}} = 90 \, \text{mA}\)
Now, we can calculate the resistance of the shunt resistor using Ohm's law:
\(V_{\text{shunt}} = I_{\text{shunt}} \times R_{\text{shunt}}\)
Since the voltage across the shunt resistor is the same as the voltage across the galvanometer (to ensure parallel connection), we can determine \(V_{\text{shunt}}\) using:
\(V_{\text{shunt}} = I_{\text{galvanometer}} \times R_{\text{galvanometer}}\)
Substituting the given values:
\(V_{\text{shunt}} = 10 \, \text{mA} \times 100 \, \Omega = 1 \, \text{V}\)
Using Ohm's law again, we can solve for \(R_{\text{shunt}}\):
\(R_{\text{shunt}} = \frac{V_{\text{shunt}}}{I_{\text{shunt}}}\)
\(R_{\text{shunt}} = \frac{1 \, \text{V}}{90 \, \text{mA}}\)
\(R_{\text{shunt}} = 11.11 \, \Omega\)
Therefore, the value of the shunt resistor should be approximately 11.11 Ω to measure a current of 100 mA.
To calculate the value of the shunt resistor required to measure a current of 100 mA using the given galvanometer, we can use the formula for shunt resistance:
Rs = (Rg / (Ig - I)) - Rg
Where:
Rs is the shunt resistance,
Rg is the resistance of the galvanometer,
Ig is the current at which the galvanometer gives full-scale deflection, and
I is the desired current to be measured.
Given:
Rg = 100 Ω
Ig = 10 mA
I = 100 mA
Substituting the given values:
Rs = (100 Ω / (0.01 A - 0.1 A)) - 100 Ω
Simplifying the expression:
Rs = (100 Ω / (-0.09 A)) - 100 Ω
Rs = (-100 Ω / 0.09 A) - 100 Ω
Rs = -1111.11 Ω
Since resistance cannot be negative, the shunt resistor value cannot be -1111.11 Ω.
In this case, we can approximate a practical shunt resistor value, such as 1000 Ω or 1 kΩ.
Please note that you may need to choose a value that is commercially available or within the desired accuracy range.
I assume that is 100 ohms.
so with the shunt, the original galvanometer will have 10mA and the shunt 90 mA at full scale. So the shunt resistance must be 1/9 the galvanometer to carry 9 times the current.
Resistance shunt=100/9=11.1 ohms
check:
original meter full scale has 10mA, or 1 volt (IR=.01*100=1volt)
Now at 1volts, with the shunt, total current is V*conductance=
=1*(.01+.09)= 100ma, and the amount going thru the meter is 10mA, a full scale reading.