That's on page 345 of your text book.

Is the question complete in itself, to lead us to an answer?

Please turn off the all-caps; the forum doesn't like shouting.

15 Mega-amps? Jee willikins! You don't half have to invest in some step-up equipment, Mildred!

For a full scale deflection of 15 milliamps [I] with resistance of 4 0hm [R], apply the standard formula for voltage V across a resistor R carrying current I, note the formula assumes units of volts, ohms and amperes :-

V = IR = I x R

Note: this comes from Ohm's Law, I = V/R, by multiplying both sides by R, hence RI = VR/R. Since R/R =1, RI = V and hence V =IR. By similar algebra, R = V/I.

For given values ( 1 milliampere (mA) = 0.001 amperes) of 15 mA and 4 ohms :-

V = 0.015 x 4 = 0.06 volts = 60 millivolts.

To get 0.06 volts at 0.1 amps across a circuit formed by the shunt in parallel with the meter resistance of 4 ohms, then the shunt must drop 0.06 volts at (0.1 - 0.015) = 0.085 amps (the meter carries the rest of the 0.1 amp).

So, applying R = V/I to the shunt of resistance r ohms,

r = 0.06/0.085 = 0.70588 ohms.

Check 0.70588 ohms x 0.085 amps = 0.0599999 volts

In practice, you will find that a 1% accurate shunt is easy to get, the price and availability of an 0.1% accurate shunt is much more challenging, but 0.01% is very rare and expensive. So 0.07059999 ohms as a check result is good enough.

Regards,

67model