A galvanometer of resistance 20 ohms is to be provided with a shunt that 1/10 of the whole current in acircuit passes through the galvanometer. The resistance of the shunt is. Please explain your answer thanks.?
what do you know about resistances in parallel?
To find the resistance of the shunt, we need to understand the concept of current division and Ohm's Law.
In this case, we have a galvanometer with a resistance of 20 ohms. We want to design a shunt that allows 1/10th of the current to pass through the galvanometer. Let's call the total current in the circuit I.
According to current division, the current passing through the galvanometer (I_g) is equal to 1/10th of the total current (I), and the current passing through the shunt (I_s) is equal to 9/10th of the total current (I).
Mathematically, we can represent it as:
I_g = (1/10) * I
I_s = (9/10) * I
Now, let's apply Ohm's Law:
Ohm's Law states that the current flowing through a resistor is directly proportional to the voltage across it and inversely proportional to the resistance of the resistor.
In this case, the voltage across the galvanometer and the shunt is the same since they are connected in parallel. Let's call this voltage V.
Using Ohm's Law, we can express the current passing through the galvanometer and shunt in terms of the voltage and resistance:
I_g = V / R_g
I_s = V / R_s
Where R_g is the resistance of the galvanometer, and R_s is the resistance of the shunt.
Now, we can equate the equations for I_g and I_s:
V / R_g = (1/10) * I
V / R_s = (9/10) * I
Since the voltage (V) and total current (I) are the same in both equations, we can equate the two equations:
V / R_g = V / R_s
Simplifying the equation, we get:
R_s = R_g * (9/10)
Plugging in the given resistance of the galvanometer (R_g = 20 ohms), we can calculate the resistance of the shunt (R_s):
R_s = 20 * (9/10)
R_s = 18 ohms
Therefore, the resistance of the shunt is 18 ohms.