An ammeter of resistance 1. 5Ω can measure currents upto 1A . The...

Given, for an ammeter, ${I}_{g} = 1A, {R}_{g} = 1 . 5Ω$

Total current, $I = 4A$

If ${R}_{s}$ be the shunt resistance, then~according to the question circuit diagram is given,& $\left({R}_{s} ‖ {R}_{g},\right) hence have equal potential difference$)$</p><p><img src="/api/proxy-image?url=https${I}_{g}{R}_{g} = \left(I - {I}_{g}\right){R}_{s}$</p><p>${R}_{s} = \dfrac{Ig}{I - {I}_{g}} \times {R}_{g} = \dfrac{1}{4 - 1} \times 1 . 5 = 0 . 5Ω$