The shunt must carry everything the movement cannot, and both share the same voltage.

Step 1 — the current split. Of the 5 A entering, the movement takes only its full-scale current and the shunt takes the rest:

\(I_{sh}=I-I_{m}=5-300\times10^{-6}=4.9997\ \text{A}\)

Step 2 — equate the voltages. The shunt is in parallel with the movement, so

\(I_{m}R_{m}=I_{sh}R_{sh}\)

\(R_{sh}=\dfrac{I_{m}R_{m}}{I-I_{m}}=\dfrac{300\times10^{-6}\times75}{4.9997}\)

\(=\dfrac{0.0225}{4.9997}=0.0045\ \Omega\)

— option 3, that is 4.5 mΩ.

A sanity check on the magnitude. The multiplying power of the shunt is

\(n=\dfrac{I}{I_{m}}=\dfrac{5}{300\times10^{-6}}=16{,}667\)

and since \(R_{sh}=R_{m}/(n-1)\), the shunt must be about 16 666 times smaller than the 75 Ω coil — roughly 4.5 mΩ. A milliohm-scale answer is exactly what a large multiplying power demands, so any option in ohms would have been wrong by orders of magnitude.

Note that subtracting I_m barely matters here. Using 5 A instead of 4.9997 A changes the answer by 0.006 %, far below the tolerance of any real shunt. The subtraction is nonetheless the correct form, and it becomes significant when the meter's full-scale current is an appreciable fraction of the range.

What makes a shunt difficult in practice. A resistance of a few milliohms is comparable with the resistance of the leads and joints connecting it, so a shunt is built as a four-terminal component: two heavy current terminals and two separate potential terminals placed inside them, so that the measured voltage is developed across a defined length of the resistive element and the contact resistances fall outside the measurement. Manganin is used for the element because its temperature coefficient is nearly zero, which keeps the calibration stable as the shunt heats up under 5 A.

Hence, the shunt resistance is 0.0045 Ω.