A galvanometer of resistance 100 Ω gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0– 10 A. The shunt required is: ____.

We are given a galvanometer that has a resistance $R$_g = 100 Ω and gives a full‑scale deflection when a current of $I$_g = 1 mA = 0.001 A flows through it.

To turn this galvanometer into an ammeter whose full‑scale range is 0 – 10 A, a shunt resistor $R$_s is connected in parallel with the galvanometer. The shunt allows most of the current to bypass the galvanometer, while only $I$_g continues to flow through the galvanometer.

The total current that the ammeter must measure at full scale is $I$ = 10 A.

Because the galvanometer and the shunt are in parallel, the voltage across both elements is the same. Therefore,

$V=I$_g·R_g = I_s·R_s

where $I$_s is the current through the shunt. The shunt current is the total current minus the galvanometer current:

$I$_s = I - I_g

Substituting the values,

$I$_s = 10 A - 0.001 A = 9.999 A

Now solve for the shunt resistance $R$_s using the equality of voltages:

$R$_s = \frac{I_g·R_g}{I - I_g}

Insert the numerical values:

$R$_s = \frac{0.001 A × 100 Ω}{9.999 A}

Calculate the numerator:

$0.001 A\; \times \; 100 \Omega \; =\; 0.1$

Finally, divide by the denominator:

$0.1\; ÷\; 9.999\; \approx \; 0.010001 \Omega$

Rounding to the precision given in the options, the required shunt resistance is approximately 0.01 Ω**.