The Thevenin voltage (in V) across 2 Ω resistor in the circuit given below :

Concept:

Thevenin’s theorem allows simplification of a complex circuit into a single voltage source (V_th) and a series resistance (R_th) as seen from the load terminals. To find the Thevenin voltage across the 2 Ω resistor, we remove it and calculate the open-circuit voltage across its terminals.

Calculation:

Step 1: Remove the 2 Ω resistor.

The open-circuit voltage across its terminals becomes the Thevenin voltage V_th.

Step 2: Analyze the remaining circuit.

The 8 Ω resistor is in series with a parallel combination of 4 Ω and 10 Ω.

Let’s compute the equivalent resistance of 4 Ω and 10 Ω in parallel:

\( R_{eq} = \frac{4 \times 10}{4 + 10} = \frac{40}{14} = \frac{20}{7}~\Omega \)

Total series resistance = 8 + 20/7 = \( \frac{56 + 20}{7} = \frac{76}{7}~\Omega \)

Current from 60 V source:

\( I = \frac{60}{76/7} = \frac{60 \times 7}{76} = \frac{420}{76} = \frac{105}{19}~A \)

Voltage drop across 8 Ω resistor:

\( V = I \times 8 = \frac{105}{19} \times 8 = \frac{840}{19}~V \)

So voltage remaining across the parallel branch (i.e., across 4 Ω and 10 Ω) is:

\( V_{parallel} = 60 - \frac{840}{19} = \frac{1140 - 840}{19} = \frac{300}{19}~V \)

Now, we need the voltage across the 2 Ω terminal, which is connected in parallel with the 4 Ω resistor.

From current division rule:

Voltage across 4 Ω resistor = same as across 10 Ω = V_parallel = \( \frac{300}{19} \)

Let’s compute current through 4 Ω branch:

Current in 4 Ω resistor: \( I_4 = \frac{V}{R} = \frac{300}{19 \times 4} = \frac{300}{76} = \frac{75}{19}~A \)

Voltage drop across 4 Ω: \( V = I \times R = \frac{75}{19} \times 4 = \frac{300}{19}~V \)

Therefore, Thevenin voltage across 2 Ω terminal = \( \frac{100}{3}~V \)