Using Kirchoff's rules, calculate the current through the $40\; \Omega$ and $20\; \Omega$ resistors in the following circuit.

• According to Kirchoff's loop rule, The sum of potential differences across a closed loop is zero.

In ABCDA, applying Kirchoff's Loop Rule -

$80-20I_{1}-40I_{2}=0$

$8-2I_{1}-4I_{2}=0$

On rearranging, the equation we have,

$I_{1}+2I_{2}-4=0$ ___(1)

In $\inline CDEFC,$ applying Kirchoff's Loop Rule -

$\inline -40I_{2}-40+10 (I_{1}-I_{2})=0$

$\inline -40\; I_{2}-40+10\; I_{1}-10 \; I_{2}=0$

$\inline -50\; I_{2}-40+10\; I_{1}=0$

On rearranging, the equation we have

$\inline I_{1}-5\; I_{2}-4=0$ _____(2)

from (1) and (2)

$\inline I_{1}-5\; I_{2}-4-I_{1}+2I_{2}-4=0$

$\inline 7\; I_{2}=0$

$\inline I_{2}=0$

On equaling the value of $\inline I_{2}$ in equation (1) , we get

$\inline I_{1}=4A$

Hence, there is no current through  the $\inline 40\Omega$ resistor and $\inline 4A$ current flows through the $\inline 20\Omega$ resistor.

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