In the given network calculate the value of the current in each resistor

Dear Student,

Please find below the solution to the asked query:

Here, the three resistors are connected in parallel. So, the current through each resistor are, here, the ratio currents through each resistance are,

$i_{4\Omega }:i_{8\Omega }:i_{16\Omega }=4:2:1$

Let the current through 16 ohm resistor is i'. Then,

$$\begin{gathered} if i_{16\Omega }=i\text{'}, then i_{4\Omega }=4i\text{'}; i_{8\Omega }=2i\text{'} \\ \therefore i_{16\Omega }+i_{8\Omega }+ i_{4\Omega }=1 A \Rightarrow i\text{'}+2i\text{'}+4i\text{'}=1A \Rightarrow 7i\text{'}=1 A \Rightarrow i\text{'}=\frac{1}{7}A \\ So, \\ {i}_{16\Omega }=\frac{1}{7}A; i_{8\Omega }=\frac{2}{7}A \& i_{4\Omega }=\frac{4}{7}A \end{gathered}$$

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