Question

For the circuit shown in the figure, the current in the 8-Ω resistor is 0.50 A, and all quantities are accurate to 2 significant figures. What is the current in the 2-Ω resistor?

Answer 1

Apply Kirchoff's voltage rule for the upper small loop to calculate the current through \(16 \Omega\).

$$ \begin{aligned} I_{16 \Omega}(16 \Omega) &=(0.5 \mathrm{~A})(8 \Omega) \\ I_{16 \Omega} &=\frac{(0.5 \mathrm{~A})(8 \Omega)}{(16 \Omega)} \\ &=0.25 \mathrm{~A} \end{aligned} $$

The total current through the resistor of \(20 \Omega\) is, \(I_{200}=0.25 \mathrm{~A}+0.5 \mathrm{~A}=0.75 \mathrm{~A}\)

The total resistance in the upper loop is,

$$ R=\frac{1}{\left(\frac{1}{16 \Omega}+\frac{1}{8 \Omega}\right)}+20 \Omega=25.3 \Omega $$

The voltage is, \(V=I R=(0.75 \mathrm{~A})(25.33 \Omega)=18.99 \mathrm{~V} \approx 19 \mathrm{~V}\)

The current passing through the resistor of \(2 \Omega\) is, \(I_{2 \Omega}=\frac{19 \mathrm{~V}}{2 \Omega}=9.5 \mathrm{~A}\)