Question

An ammeter of resistance $0.80\Omega$ can measure current up to 1.0A.(i) What must be the value of shunt resistance to enable the ammeter to measure current up to 5.0A.
(iii) What is the combined resistance of the ammeter and the shunt?

Answer 1

(i) The net resistance of ammeter with shunt connected in parallel=$R_2=\frac{R_{A}x}{R_A+x}$

$=\frac{0.8\times x}{0.8+x}$

The potential of the cell is

$V=I_1{R}_A=I_2{R}_2$

$⟹1\times 0.8=5\times \frac{0.8x}{0.8+x}$

$⟹x=0.2\Omega$

(ii) Combined resistance of the ammeter and thus shunt

$R=\frac{0.8x}{0.8+x}=0.16\Omega$