As shown in the figure, a 1Ω resistance is connected across a source that has a load line V + i = 100. The current through the resistance is

Concept:

Thevenin's Theorem:

Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.

To find V_oc: Calculate the open-circuit voltage across load terminals. This open-circuit voltage is called Thevenin’s voltage (V_th).

To find I_sc: Short the load terminals and then calculate the current flowing through it. This current is called Norton current (or) short circuit current (i_sc).

To find R_th: Since there are Independent sources in the circuit, we can’t find Rth directly. We will calculate Rth using Voc and Isc and it is given by

\({{\rm{R}}_{{\rm{th}}}} = \frac{{{{\rm{V}}_{{\rm{oc}}}}}}{{{{\rm{i}}_{{\rm{sc}}}}}}\)  

Application:

Given: Load line equation = V + i = 100

To obtain open-circuit voltage (Vth) put i = 0 in load line equation 

⇒ Vth = 100 V

To obtain short-circuit current (isc) put V = 0 in load line equation

⇒ isc = 100 A

So, \({R_{th}} = \frac{{{V_{th}}}}{{{i_{sc}}}} = \frac{{100}}{{100}} = 1{\rm{\Omega }}\)

Equivalent circuit is

Current (i) = 100/2 = 50 A

Applying loop-law in the given circuit.

- V + i × R = 0

- V + I × 1 = 0

⇒ V = i

Given Load line equation is V + i = 100

Putting V = i 

then i + i = 100 

⇒ i = 50 A