Find the current flowing through 10 Ω load resistors shown in the given circuit, using Thevenin's theorem or Norton's theorem.  

The correct answer is option 4):(0.32 Amps)

Concept:

According to Thevenin’s theorem, any linear circuit across a load can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth as shown

V_th = Open circuit Voltage at a – b (by removing the load), i.e.

If a linear circuit contains dependent sources only, i.e., there is no independent source present in the network, then the open-circuit voltage or Thevenin voltage will simply be zero. (Since there is no excitation present)

Calculation:

The given circuit is 

Step 1:To find V_Th remove the 10 Ω resistor

The 1 Ω resistor and 4 Ω form a branch

The current through this branch is

I₁ = \(10 \over 5\) = 2 A

 The 5 Ω  and 3 Ω form a branch

The current through the branch is

I₂ = \(10 \over 8\)

= 1.25 A

The Thevenin voltage will be

V_th = Voltage across the 10 Ω  = 4I₁ - 3 I2

= 4.25 V

Step 2:To find RTh Open circuit the 10 Ω resistor branch

Then 

(4 Ω || 3 Ω) series with (5 Ω || 1 Ω)

R_th = ( \(1 \over 4\) + \(\frac{1}{3}\) )^-1 + ( \(1 \over 5\) + \(1 \over 1\) )^-1

= 1.71 + 0.833

= 2.54Ω

Step 3: Thevenin equivalent ckt

I_L  is the current through a 10 Ω resistor

IL = \(V_{Th} \over (R_{Th} + R_{L})\)

R_L = 10 Ω

I_L = \(4.25 \over 10+ 2.54\)

= 0.32 A