من آر تونن را نیم اهم و آی نورتن را ۴ آمپر و ولتاژ...

생성일: 2025년 11월 7일

답변에 사용된 모델 GPT-5 Thinking by Chat01

질문

Screenshot_۲۰۲۵۱۱۰۷_۱۲۳۷۰۹_Drive.jpg

من آر تونن را نیم اهم و آی نورتن را ۴ آمپر و ولتاژ تونن را ۲ ولت آوردم
آیا این برای این سوال درسته

사고

Analyzing the circuit

There’s a 20A current source at the left of the resistor, flowing downward from top to bottom. There’s also a vertical short between top and bottom, with an upward current (i1). A 2V controlled voltage source is in series, dependent on the output port voltage. We need to verify Thevenin and Norton values, such as R_th=0.5Ω, I_th=4A, and V_th=2V. Let's dig into this analysis to ensure the relationships hold.

Analyzing Node N1

The mid-point node between the resistor and dependent elements is called N1, which is the top of the dependent current source too. The voltage of the dependent voltage source is 2V, and the potential difference between nodes N1 and a is 2V. Simplifying the expression, we find V(N1) = 3V. Now, we can use Kirchhoff's Current Law (KCL) to analyze the currents leaving node N1.

Defining currents and KCL

The dependent voltage source is in series but has an unspecified current. I'll define the current, "i," entering the network at terminal a, flowing leftwards through the voltage source. Current through the dependent voltage source is i directed from right to left, while at node N1, the current is -i. Using Kirchhoff's Current Law (KCL), I'll treat these currents as unknowns and formalize the solution. Additionally, KCL at node N0 considers the 20A source, the short branch current, and the resistor.

Defining the nodes and elements

I have four key nodes to define:

Ground node (b) at 0V.
Node A, located to the right of the dependent voltage source, where V(a) = v.
Node 1, between the dependent voltage source and the resistor, where V(N1) = 3V.
Node 0 at the top left, near the 20A current source and vertical short, where I’ll call V0.

The 20A current source injects current downward from node 0 to ground.

Analyzing KCL and currents

At node 0, we have several components: The 20A current source, the short circuit, and the resistor through which current flows differently. For KCL at node 0, the sum of the currents leaving should equal 0. After analyzing, I conclude that V0 equals 0, as node 0 is tied to ground, simplifying the network. The current-related equations follow from there and depend heavily on the resistor and short branch.

Clarifying current behavior

It looks like the current source is shorted since both top and bottom are the same node, meaning there’s no net injection at any point. The problem is that the current through the vertical short isn’t well-defined in circuit analysis—it depends on how we interpret the current through the short. So, I need more precision regarding current distribution across branches.

Solving for i1

I’m thinking that the current through the short branch might be undefined, but if we look at KCL at the top node, we can derive i1 through the equation: i1 = 20 - V1. This solution seems more reasonable now that we have the proper current relationships between branches, even with the bottom being the same node.

Troubleshooting circuit analysis

Looking at the KCL at the top node, I derived i1 = 20 - V1, which fits, while also confirming v = V1/3. However, the confusion arises with the Norton current and Rth. The calculated Norton current is 20A, but the user claimed 4A, which is inconsistent. The calculated Rth is 0.1667Ω, not 0.5Ω as the user mentioned.

Clarifying Thevenin resistance and Norton current

The discrepancies likely stem from the different methods I've used to calculate the Thevenin resistance and Norton current. While I get 20A for the Norton current, the user's value of 4A fits with their Rth of 0.5Ω. The key difference appears to be in assumptions or interpretations of certain elements, like the voltage sources and short circuits.

Revisiting circuit relationships

I believe the key relationships are now clearer. From KCL and the earlier results, I have consistent outcomes: I found that i1 = 20 - V1 holds true. The KCL method confirmed V_th ≈ 3.33V and R_th ≈ 0.1667Ω. The breakdown of current through various elements checks out, but I'm still double-checking the direction conventions.

Thought for 4m 57s

답변

نه—این اعداد درست درنمیاد.

خلاصهٔ حل (خیلی کوتاه):

ولتاژ دو سر المان وابسته: $V_1 - v = 2v \Rightarrow V_1=3v$ .
در حالت مدار باز ( $i=0$ ): سوپرنود می‌دهد $V_1-i_1=0$ و از KCL روی شینِ چپ $i_1+V_1=20$ . پس $V_1=10\Rightarrow v=V_1/3=10/3\approx3.33\text{ V}$ .
$\Rightarrow V_{\text{Th}}=10/3\text{ V}$
در اتصال‌کوتاه ( $v=0$ ): منبع ولتاژ وابسته صفر ولت می‌شود $\Rightarrow V_1=0$ . از KCL: $i_1=20$ و $i=V_1-i_1=-20\text{ A}$ . بنابراین جریان نورتن (از شبکه به بیرون) $I_{\text{N}}=20\text{ A}$ است.
مقاومت تونن: $R_{\text{Th}}=\dfrac{V_{\text{Th}}}{I_{\text{N}}}=\dfrac{10/3}{20}=\dfrac{1}{6}\approx0.167\ \Omega$ .

یک رابطهٔ کلی بین ترمینال‌ها هم درمی‌آید:

i=6v-20 \quad\Rightarrow\quad v=\frac{1}{6}i+\frac{10}{3}

(جهت $i$ طبق شکل به سمت چپ و داخل مدار است؛ به همین خاطر در اتصال‌کوتاه $i=-20$ A ولی $I_{\text{N}}=+20$ A از a به b است.)

پس مقادیر درست:

V_{\text{Th}}=\tfrac{10}{3}\text{ V},\quad I_{\text{N}}=20\text{ A},\quad R_{\text{Th}}=\tfrac{1}{6}\ \Omega

نه $2\text{ V}, 4\text{ A}, 0.5\ \Omega$ .

질문

من الان منابع مستقل حذف کردم و بعد یک منبع جریان ۱ آمپری برای تست گذاشتم که ولتاژ وی داره بعد آر تونن را که حساب می کنم مساوی وی میشه را نیم اهم به دست میارم کجای کارم اشتباه توی این کار منبع جریان ۲۰ آمپری و منبع ولتاژ ۲ ولتی حذف می شود و حساب کتاب انجام میشه همین طور بعدش isc را حساب می کنم و بعدش میام از روی Rth و Isc میایم Vth را حساب می کنم که ۲ ولت میشه اشتباهم کجاست و حتی برای isc یک اتصال کوتاه رح میده و منبع جریان ۲۰ آمپری حذف میشه

사고

Clarifying the circuit

It looks like there's a current source (i1) located mid-right between the rails and the node with the resistor and VCVS. The drawing shows an upward arrow labeled i1, indicating current flows through this path. The dependent source injects the same current (i1) at a node between the resistor and VCVS.

Considering dependent sources and ISC

The circuit has a dependent current source controlled by i1, which comes from a short-circuit branch. Even with independent sources deactivated, the short may carry a test current. Since Isc involves all sources being active, the statement about removing the current source for Isc is incorrect. All sources must remain active when calculating Isc.

Evaluating Rth calculation

The user seems to have removed both the 20A source and the 2V source (which is actually dependent, not independent) and added a 1A test source to compute Rth. In standard Rth computation, dependent sources must remain active while independent ones are zeroed. Let’s use KCL with the test source set up and evaluate why their Rth value of 0.5 ohm is incorrect.

Analyzing circuit nodes

The VCVS depends on the terminal voltage, and it's in series with a resistor forming a network. At the bottom, we designate node b as ground. Node L is tied to the top of the 20A source and connects to the left short (i1) that leads to ground. This short makes Node L effectively ground. Hence, V1, the voltage between the resistor, current source, and VCVS, is measured relative to ground.

Analyzing the circuit with deactivated sources

The 20A source deactivates, and the vertical short creates a direct connection between top and bottom, making i1 zero. In this state, the dependent current source goes off (i1=0). With the left branch open, there's no current through the short. We can compute Rth with the test current and dependent VCVS, which imposes a relation: V1 = 3v. Without a current source, the KCL at V1 will involve the voltage source and resistor to ground.

Nodal analysis of test current

In the current setup, we are using a test source in the circuit. The VCVS defines a relationship between node V1 and node a, representing the supernode. Using KCL, we deduce that the current through the 1Ω resistor must equal the test current. This results in V1 = 1V, and consequently, v = 1/3 V. However, the earlier method gave Rth as 1/6 Ω, suggesting a discrepancy caused by assumptions about the dependent current source.

Re-evaluating the analysis

I need to reconsider the current source calculations. My earlier calculation for Rth was based on a 20 A current source, but that might have led to a sign issue. When re-evaluating Vth in an open-circuit condition, I realize I need to carefully apply KCL at each node, especially considering that the currents from the VCVS and the short branches affect the total current distribution. I'll continue to refine this analysis.

Reaffirming the calculations

I’ve worked through the KCL equations and confirmed that V1 = 10 V and v = 10/3 V. Now, with the short-circuit scenario, V1 = 0 and the current through the resistor is zero. The dependent current source i1 is 20A, as confirmed by the KCL at the top-left node.

Clarifying node relations and currents

The dependent current source at node V1 injects current i1 upward, flowing through the VCVS and the short to ground, which equals 20A when the short-circuit occurs. Now, with the independent 20A source removed, i1 still exists due to the dependency on the VCVS, and we need KCL to properly express this relationship. Hence, we continue solving for i1 and the rest of the variables.

Testing Rth with conditions

When we remove the 20A current source, we establish i1 = -V1, resulting in the correct calculations for Rth. Without turning off dependent sources, KCL at the top-left node shows how i1 behaves as it remains tied to V1. The confusion earlier stemmed from incorrectly setting i1 = 0, which misled the Rth value.

Thought for 4m 21s

답변

مشکل از دو جاست:

برای محاسبه‌ی $R_{\text{Th}}$ باید فقط منابع مستقل را صفر کنید: منبع جریان $20\text{A}$ → مدار باز؛ (اگر منبع ولتاژ مستقلی بود) → اتصال‌کوتاه. منابع وابسته (منبع ولتاژ $2v$ و منبع جریان وابسته به $i_1$ ) باید حتماً بمانند. شما عملاً منبع $2v$ را هم «حذف» کرده‌اید و $i_1$ را صفر گرفته‌اید، که درست نیست.
وقتی منبع $20\text{A}$ را باز می‌کنید، هنوز جریان $i_1$ از قانون KCL گره چپ به دست می‌آید و صفر نیست. در آن گره فقط مقاومت $1\Omega$ و همان شنت وجود دارند، پس:

i_1+V_1=0\quad\Rightarrow\quad i_1=-V_1

روش صحیح با منبع آزمون (برای $R_{\text{Th}}$ )

یک منبع جریان آزمون $1\text{A}$ از $a$ به $b$ بگذارید (به داخل مدار).
منبع وابسته‌ی ولتاژ الزام می‌دهد: $V_1-v=2v \Rightarrow V_1=3v$ .
KCL روی سوپرنود (گره‌های $V_1$ و $v$ ):
جریان خروجی به زمین از سوپرنود = جریان منبع آزمون $\frac{V_1}{1}-i_1=1$
با $i_1=-V_1$ : $V_1-(-V_1)=1 \Rightarrow 2V_1=1 \Rightarrow V_1=\tfrac{1}{2}\text{ V}$
پس $v=V_1/3=\tfrac{1}{6}\text{ V}$ و

R_{\text{Th}}=\frac{v}{1\text{A}}=\boxed{\tfrac{1}{6}\ \Omega}

برای $I_{\text{sc}}$ (نورتن)

اینجا هیچ منبعی را حذف نکنید؛ فقط پایانه‌ها را اتصال‌کوتاه کنید: $v=0\Rightarrow V_1=0$ . از KCL گره چپ:

i_1+V_1=20 \Rightarrow i_1=20\text{ A}

بنابراین جریان اتصال‌کوتاه (از مدار به شنت) $=20\text{A}$ است؛ با جهت پیکان شکل، $i_{\text{sc}}=-20\text{A}$ ولی مقدار نورتن $\boxed{I_{\text{N}}=20\text{ A}}$ .

و در نتیجه

V_{\text{Th}}=R_{\text{Th}}I_{\text{N}}=\tfrac{1}{6}\times 20=\boxed{\tfrac{10}{3}\text{ V}}

پس اعداد درست: $R_{\text{Th}}=\tfrac{1}{6}\Omega$ ، $I_{\text{N}}=20\text{A}$ ، $V_{\text{Th}}=\tfrac{10}{3}\text{V}$ . اشتباه شما این بود که منبع وابسته را خاموش کردید و $i_1$ را صفر فرض کردید، و همچنین برای $I_{\text{sc}}$ منبع $20\text{A}$ را حذف کردید.

질문

Screenshot_۲۰۲۵۱۱۰۷_۱۲۳۷۱۴_Drive.jpg

می تونی این سوال هم گام به گام و کامل توضیح بدهی و تا آخر حل کنی

사고

Guiding step-by-step

In study mode, I can't just give the final answer right away. I need to guide you through a structured process. Since you're asking for step-by-step help, I'll ask you questions one at a time.

Let’s start with understanding the circuit. I see there’s a 4V source connected to a 7Ω and 2Ω resistor. How can we break down this circuit using the principle of superposition?

Clarifying the circuit setup

Okay, let’s outline the components. We have a 7Ω resistor from the left top node to the central top node, and then a 2Ω resistor down to the ground. There’s also a dependent current source (0.4 * i1) on the right branch.

Now, for the top right nodes—node A and node B—there's a 6A current source and a 1Ω resistor in parallel between them.

How does this setup help us apply the superposition principle?

Simplifying the setup

We have two independent sources: a 4V voltage source on the left and a 6A current source on the top right. There’s also a dependent current source (0.4 * i1) that depends on the current i1 through the 2Ω resistor. To solve it, we'll use the superposition principle, considering each source separately.

Next, we’ll focus on the 4V voltage source. How would you approach this?

Considering each case

Under superposition, we keep dependent sources active, but we turn off independent ones. So for Case 1, we’ll keep the 4V source active and open the 6A source (which becomes an open circuit). The current i1 through the 2Ω resistor is related to the voltage at node A. Let’s analyze this further—what does the voltage at node A look like now?

Guiding through the steps

For Case 1, we need to break down the equations step-by-step. In the nodal equation at node A, we use the KCL equation, considering the resistances and the dependent current source. I want to guide you to:

Set VA = ... and VB = ...
Use i1 = VA / 2.
Develop the equations from KCL.

Let’s start with the first step! Can you try applying KCL to node A?

Clarifying KCL for supernode

For Case 2, we want to analyze the currents at nodes A and B using a supernode approach. This requires writing the KCL considering all currents leaving the supernode, including the 7Ω resistor from A to ground, 2Ω from A, 3Ω from B, and the dependent current source.

By treating both nodes A and B as one supernode, we'll simplify the process of evaluating the current contributions! Let's proceed with this approach—can you try setting up the equation based on this setup?

Deriving KCL for Node A

At node A, we have the currents leaving through the 7Ω, 2Ω resistors, 1Ω, and the current source. Using KCL, the sum of the currents leaving should equal zero. We treat the current source as +6A leaving node A. This leads to the equation:

(VA /7) + ((VA - VB)/1) + (VA/2) + 6 = 0.

Now, the question is whether this fits physically, considering the signs involved and current directions.

Confirming next steps

To solve this step-by-step, first, we need to calculate VB1 through the components we’ve derived. The goal is to reach v. Let’s start with this first stage and double-check if you're comfortable with it. Are you ready to proceed? I would like to keep it minimal, so let me know!

Thought for 2m 57s

답변

عالیه—با هم «گام‌به‌گام» و با جمع آثار می‌ریم جلو. هدف: ولتاژ $v$ روی مقاومت $3\Omega$ (همون $V_B$ ).

نقشهٔ راه

دو منبع مستقل داریم: منبع ولتاژ $4\text{V}$ و منبع جریان $6\text{A}$ .
در هر حالت، منبع وابسته $0.4 i_1$ را خاموش نمی‌کنیم.
گره‌ها را این‌طور نام می‌ذاریم: $V_A$ گره وسط بالا، $V_B$ گره راستِ بالا (همون $v$ ).
جریان مرجع $i_1$ طبق شکل رو به پایین از مقاومت $2\Omega$ است، پس

i_1=\frac{V_A}{2}.

حالت ۱: فقط منبع ۴V فعّال (منبع ۶A را «بازمدار» کن)

رابطهٔ منبع وابسته: جریان رو به بالا $=0.4i_1=0.2V_A$ .

بنویس KCL:

گره $A$ :

\frac{V_A-4}{7}+\frac{V_A-V_B}{1}+\frac{V_A}{2}=0 \quad\Rightarrow\quad \boxed{23V_A-14V_B=8}

(بعد از ضرب طرفین در ۱۴)

گره $B$ :

\frac{V_B-V_A}{1}+\frac{V_B}{3}=0.2V_A \quad\Rightarrow\quad \boxed{4V_B-3.6V_A=0}

(یعنی $V_B=0.9V_A$ )

سؤال ۱: از این دو معادله $V_A$ و سپس سهم ولتاژ $v_1=V_B$ را حساب کن. به چه عددی می‌رسی؟
(فقط نتیجهٔ $v_1$ را بگو، بعد می‌ریم حالت ۲.)

질문

사고