Question

Find the Norton equivalent circuit across A-B terminal for the circuit shown

Answer 1

Hint: The Norton current I is the open circuit voltage divided by the resistance r across that terminal. The Norton resistance r across A-B with all voltage sources replaced by short circuit and all current sources replaced by open sources.

Complete step by step answer:

Step 1: We have to find the load resistance. Load resistance is the resistance through which we have to find the Norton current. So here we have to find the current through A-B terminal so the load resistance is 10Ω.

$R_L$ = 10 Ω

Step 2: We will find Norton equivalent resistance ($R_n$) which will be equivalent to the resistance across A-B terminal. For this we will short circuit the voltage source and open circuit all the current source  as shown in the figure. Equivalent resistance across the A-B terminal is $R_n$.

5 Ω and 10 Ω are in parallel with each other. Their equivalent is $R$,

$\dfrac{1}{R} = \dfrac{1}{5} + \dfrac{1}{{10}}  $

$\Rightarrow \dfrac{1}{R} = \dfrac{{5 + 10}}{{10 \times 5}}  $

$\Rightarrow \dfrac{1}{R} = \dfrac{{15}}{{50}}  $

$\Rightarrow R = \dfrac{10}{3} $

Now this $R$ is in series with 3 Ω.

$R_n = \dfrac{{10}}{3}\Omega  + 3\Omega    $

$\Rightarrow  R_n =  \dfrac{{10 + 9}}{3}\Omega$

$\Rightarrow R_n = \dfrac{{19}}{{3}}\Omega   $ 

STEP 3: Now we will find the Norton current which is equal to $I_n$.Norton current is the current which passes through A-B terminal 

Both the 3 Ω resistance are parallel to each other so their effective resistance is 

$\dfrac{1}{R} = \dfrac{1}{3} + \dfrac{1}{3}  $

$\Rightarrow \dfrac{1}{R} = \dfrac{2}{3}   $

$\Rightarrow R = \dfrac{3}{2} $

Now $R$ is in series with 5 Ω 

$R = \dfrac{3}{2} + 5   $

$\Rightarrow R = \dfrac{13}{2}$

$\Rightarrow \text{CURRENT} = \dfrac{\text{VOLTAGE}}{\text{RESISTANCE}}  $

$\Rightarrow I = \dfrac{{30}}{{\dfrac{{13}}{2}}} $

$\Rightarrow I = \dfrac{{30 \times 2}}{{13}}   $

$\Rightarrow I = \dfrac{{60}}{{13}} $

Now this current will be equally 

$I_n = \dfrac{{30}}{{13}}$

STEP 4: We will draw an equivalent circuit with all three $I_n , R_n$ and $R_l$ in parallel to each other.

$x = \dfrac{{R_n}}{{R_n + Rl}} \times I_n  $

$\Rightarrow x = \dfrac{{\dfrac{{19}}{3}}}{{\dfrac{{19}}{3} + 10}} \times \dfrac{{30}}{{13}} $

$\therefore x = \dfrac{{19 \times 30}}{{49 \times 13}}A $

Note:
Norton and Thevenin theorems are used to solve the complex circuit questions. Thevenin's theorem can be applied while analyzing a circuit with dependent sources. In this case, all independent sources are turned off and the $R_{Th}$ is calculated by applying a current source or voltage source at the open terminal of the circuit. When using a voltage source, it can be assumed to be 1V for easy calculations.