You are provided with a moving coil galvanometer of a coil resistance 20ohms and can carry a maximum current of 15ma .how will you design this galvanometer to register 10ma full scale deflection ?

To design the galvanometer to register 10mA full scale deflection, we need to calculate the required resistance to be connected in series with the coil of the galvanometer.

Given:
Coil resistance of the galvanometer (Rg) = 20 ohms
Maximum current the galvanometer can carry (Imax) = 15mA
Full scale deflection current required (Ifsd) = 10mA

Let the additional resistance to be connected in series with the galvanometer coil be R.

The total resistance in the circuit (Rtotal) = Rg + R

According to Ohm's Law, the total current passing through the circuit when the galvanometer is at full scale deflection should be equal to the full scale deflection current:
Ifsd = V / Rtotal

Since the galvanometer is at full scale deflection, the voltage across the galvanometer coil is equal to the maximum voltage that can be applied to the galvanometer:
V = Imax * (Rg + R)

Substitute the values into the equations:
10mA = (15mA) * (20 ohms + R) / (20 ohms + R)

Solving for R:
10mA * (20 ohms + R) = 15mA * (20 ohms + R)
200mA * ohms + 10mA * R = 300mA * ohms + 15mA * R
10mA * R - 15mA * R = 300mA * ohms - 200mA * ohms
-5mA * R = 100mA * ohms
R = -100mA * ohms / 5mA
R = 20 ohms

Therefore, the additional resistance (R) to be connected in series with the galvanometer coil to register 10mA full scale deflection is 20 ohms.