Question

Al the wires in FIGURE Q27.6 are made of the same material and have the same diameter. Rank in order, from largest to smallest, the currents $I_{\mathrm{a}}$to $I_{\mathrm{d}}$. Explain.

Short answer

$I_a=I_d>I_b=I_c$. Conservation of current insures that the current into the top junction point, $I_a$, must equal the current out of that point $I_b+I_c$. Similarly at the bottom junction point $I_b+I_c=I_d$. Since the wires are identical, $I_b=I_c$.

Step-by-step solution

Concept Introduction

As Kirchhoff's current law states, a node's total current will never exceed zero. According to this law, electric charges are conserved.

Explanation

The material of the wires and the diameter are all identical as shown in the above figure.

According to Kirchhoff's rules, the current $I_a$is equally divided into two equal parts.

Let $I_a=I$, then at the point $A$the current $I$is divided equally as $\frac{I}{2}$.

So, $I_b=I_c=\frac{I}{2}$

And at the point B, the two currents $I_b$and $I_c$are combined together and flows as$I_d.$

So the value of$I_d=I$
So finally we have $I_a=I_d=I$and $I_b=I_c=\frac{I}{2}$

Final Answer

Hence, the rank order is$I_a=I_d>I_b=I_c.$