In the circuit shown here, at which point is the current least?

Total Voltage $=(1.5+1.5) V$

$=3 \mathrm{~V}$

Total Resistance $=\left(\frac{1}{6}+\frac{1}{4}\right)^{-1}+10$

$\begin{array}{l} =\left(\frac{2+3}{12}\right)^{-1}+10 \\ =\frac{12}{5}+10 \\ =\frac{62}{5} \Omega \end{array}$

$\begin{aligned} V_{6} & =\frac{\left(\frac{1}{6}+\frac{1}{4}\right)^{-1}}{\frac{62}{5}} \times 3 \\ & =\frac{12}{5} \times \frac{5}{62} \times 3 \\ & =0.58 \mathrm{~V} \\ I_{Q} & =\frac{0.58}{6}=0.096 \mathrm{~A} \\ I_{R} & =\frac{0.58}{4}=0.145 \mathrm{~A} \\ I_{S} & =\frac{(3-0.58)}{10}=0.242 \mathrm{~A} \end{aligned}$

Here, $\ I_{Q} < I_{R} < I_{S}$