Question

What is current through the $5\Omega$ resistor?

Answer 1

Here we have a closed-loop of a conductor that we can divide into two elementary loops to simplify the problem. Also, we can use one of Kirchhoff’s laws to figure out our desired answer. Once we learn physics here, then it is a fundamental problem of algebra.

Complete step by step solution:
We take the closed-loop of current and part them into two sections $A$ and $B$ as shown in the figure below-

In both the current-carrying loop, the current is flowing counter-clockwise. So here we use Kirchhoff’s Voltage Law (KVL) to find out the current through the $5\Omega$ resistor.
Kirchhoff’s Second Law or Kirchhoff’s Voltage Law states that the algebraic sum of all the voltages around any closed loop in that circuit equals zero for a series of the closed-loop path.
We assume that the current through the $5\Omega$ resistor is $i$ (ampere).
According to Kirchhoff’s Voltage Law in the loop $A$-
$\Rightarrow 10 - 5i = 0$ ………$(1)$
Since the voltage difference across the resistor $10V$ and $i$ current is flowing through the resistor.
According to Kirchhoff’s Voltage Law in the loop $B$-
$\Rightarrow 10 - 5i = 0$ ………$(2)$
Since here also the voltage difference across the resistor $10V$ and $i$ current is flowing through the resistor.
So from both the above two equations, we can find out the value of $i$-
Hence,
$\Rightarrow 5i = 10$
$\Rightarrow i = \dfrac{{10}}{5}$
$\Rightarrow i = 2A$

Therefore, the current through the $5\Omega$ resistor is $2A$.

Note: We can use Kirchhoff’s Second Law, i.e., Kirchhoff’s Current Law, to determine the current at a junction in a closed current-carrying loop. A point of caution, we can only use these laws in a closed loop. These laws are advantageous in complex electric circuits.