Understanding the relationship between Norton and Thevenin equivalent circuits is fundamental in electrical circuit analysis. Both are simplified representations of a more complex linear circuit, but they use different approaches to model the behavior at a pair of terminals.

Norton Equivalent Circuit Basics

A Norton equivalent circuit simplifies any linear electrical network into an ideal current source ($I_N$) in parallel with an equivalent resistance ($R_N$).

• $I_N$: This represents the short-circuit current that would flow if a short were placed across the terminals where the equivalent circuit is being determined.
• $R_N$: This is the equivalent resistance looking back into the circuit from the terminals, with all independent voltage sources short-circuited and all independent current sources open-circuited. It is sometimes referred to as Norton resistance.

In this specific problem, we are given a Norton equivalent circuit with the following parameters:

• Norton Current source ($I_N$) = 100 μA
• Norton Resistance ($R_N$) = 10 kΩ

Thevenin Equivalent Circuit Basics

A Thevenin equivalent circuit, on the other hand, simplifies a linear electrical network into an ideal voltage source ($V_{Th}$) in series with an equivalent resistance ($R_{Th}$).

• $V_{Th}$: This represents the open-circuit voltage that would appear across the terminals where the equivalent circuit is being determined. It is often referred to as Thevenin voltage.
• $R_{Th}$: This is the equivalent resistance looking back into the circuit from the terminals, calculated in the same way as the Norton resistance ($R_N$).

Converting Norton to Thevenin Equivalent

The conversion between a Norton equivalent circuit and a Thevenin equivalent circuit is a common task in circuit analysis, as they are interchangeable representations of the same underlying circuit's terminal characteristics. The key relationships used for this conversion are derived from Ohm's Law and the principles of source transformation.

Since the question asks for the Thevenin voltage ($V_{Th}$), we will use the second relationship.

Calculating Thevenin Voltage ($V_{Th}$)

To determine the Thevenin voltage ($V_{Th}$) from the given Norton equivalent circuit, we apply the conversion formula $V_{Th} = I_N \times R_N$.

Let's list the given values and their standard units:

• Norton Current ($I_N$) = 100 μA (microamperes) = $100 \times 10^{-6}$ A (amperes)
• Norton Resistance ($R_N$) = 10 kΩ (kilo-ohms) = $10 \times 10^3$ Ω (ohms)

Now, substitute these values into the formula for $V_{Th}$:

$$ V_{Th} = I_N \times R_N $$ $$ V_{Th} = (100 \times 10^{-6} \text{ A}) \times (10 \times 10^3 \text{ Ω}) $$

To simplify the calculation, we can multiply the numerical parts and the powers of 10 separately:

$$ V_{Th} = (100 \times 10) \times (10^{-6} \times 10^3) \text{ V} $$ $$ V_{Th} = 1000 \times 10^{(-6+3)} \text{ V} $$ $$ V_{Th} = 1000 \times 10^{-3} \text{ V} $$

Since $10^{-3}$ means dividing by 1000:

$$ V_{Th} = 1 \text{ V} $$

Therefore, the Thevenin voltage ($V_{Th}$) for the equivalent circuit is 1 V.