Use Thevenin equivalent circuits — Hinglish notes

1.2.9 · Hardware › Circuit Analysis Fundamentals

Thevenin equivalent KYA hota hai?

YEH KAAM KYO KARTA HAI? Kyunki network linear hai, terminal voltage $V$ aur terminal current $I$ ek seedhi line ka relationship follow karte hain: $V = V_{t h} - I R_{t h}$ Ek straight line ko sirf do parameters chahiye: ek intercept ( $V_{t h}$ ) aur ek slope ( $- R_{t h}$ ). Yahi sab kuch hai jo bahar ki duniya kabhi measure kar sakti hai — isliye equivalent ko bhi bas yahi chahiye.

$V = V_{t h} - I R_{t h}$ scratch se KAISE derive karein

Yeh step kyun? Superposition sirf isliye legal hai kyunki circuit linear hai — yahi ek assumption hai jis par poora theorem tika hai.

Do numbers KAISE find karein (recipe)

$V_{t h}$ — load disconnect karo. Kisi bhi method se (node/mesh/divider) terminals ke across open-circuit voltage compute karo.
$R_{t h}$ —
- Koi dependent sources nahi? Sab independent sources deactivate karo (V-source → wire, I-source → gap) aur andar dekhte hue equivalent resistance compute karo.
- Dependent sources present hain? Unhe deactivate nahi kar sakte. Iske badle ek test source $V_{t es t}$ lagao, $I_{t es t}$ find karo, aur $R_{t h} = V_{t es t} / I_{t es t}$ lo.
Redraw karo: $V_{t h}$ series mein $R_{t h}$ ke saath, phir load reconnect karo.

Figure — Use Thevenin equivalent circuits

Worked Example 1 — ek divider with a load

Source $V_{S} = 12 V$ , $R_{1} = 4 Ω$ (series), $R_{2} = 2 Ω$ (to ground) terminals ke across $R_{2}$ banate hain. Thevenin equivalent find karo, phir load $R_{L} = 6 Ω$ mein current.

Step 1 — $V_{t h}$ (open circuit). $V_{t h} = V_{S} \frac{R _{2}}{R _{1} + R _{2}} = 12 \cdot \frac{2}{6} = 4 V$ Kyun? Load remove hone par, $R_{1}$ aur $R_{2}$ ek plain voltage divider banate hain.

Step 2 — $R_{t h}$ . $V_{S}$ kill karo (short karo). Tab $R_{1}$ aur $R_{2}$ terminals se dekhne par parallel mein hain: $R_{t h} = R_{1} ∥ R_{2} = \frac{4 \cdot 2}{4 + 2} = 1.33 Ω$ Kyun parallel? $V_{S}$ short karne se $R_{1}$ ka top ground se jud jaata hai, toh dono resistors ab terminal node se ground tak jaate hain — yahi parallel configuration hai.

Step 3 — load reconnect karo. $I_{L} = \frac{V _{t h}}{R _{t h} + R _{L}} = \frac{4}{1.33 + 6} = 0.545 A$

Worked Example 2 — short-circuit current se verify karo

Example 1 ke original circuit mein output terminals short karo. Short, $R_{2}$ ko bypass kar deta hai, toh $V_{S}$ se current sirf $R_{1}$ ke through flow karta hai: $I_{sc} = \frac{V _{S}}{R _{1}} = \frac{12}{4} = 3 A$ Check: $R_{t h} = \frac{V _{t h}}{I _{sc}} = \frac{4}{3} = 1.33 Ω.$ ✅ Step 2 se match karta hai.

Yeh kyun matter karta hai: do independent methods agree kar rahe hain → confidence ki equivalent sahi hai.

Worked Example 3 — maximum power transfer

Example 1 ke equivalent se, kaunsa $R_{L}$ sabse zyada power leta hai? Delivered power: $P_{L} = I_{L}^{2} R_{L} = (\frac{V _{t h}}{R _{t h} + R _{L}})^{2} R_{L}$ Differentiate karo aur $d P_{L} / d R_{L} = 0$ set karo: $\frac{d}{d R _{L}} \frac{R _{L}}{( R _{t h} + R _{L} ) ^{2}} = \frac{( R _{t h} + R _{L} ) ^{2} - R _{L} \cdot 2 ( R _{t h} + R _{L} )}{( R _{t h} + R _{L} ) ^{4}} = 0$ $\Rightarrow R_{t h} + R_{L} - 2 R_{L} = 0 \Rightarrow R_{L} = R_{t h}$ Toh max power $R_{L} = 1.33 Ω$ par milti hai, jisse $P_{ma x} = \frac{V _{t h}^{2}}{4 R _{t h}} = \frac{16}{4 ( 1.33 )} = 3 W$ .

$R_{L} = R_{t h}$ kyun? Thevenin poore circuit ko ek $R_{t h}$ mein collapse kar deta hai; load ko usse match karna classic impedance-matching result hai.

Common mistake Steel-manned common errors

1. " $R_{t h}$ find karte waqt dependent sources bhi deactivate karo." Kyun sahi lagta hai: Hum independent sources deactivate karte hain, toh sab sources kyun nahi? Fix: Dependent sources circuit variables par depend karte hain — unhe kill karne se circuit ka behavior badal jaata hai. Iske badle test-source method use karo.

2. " $V_{t h}$ source voltage hoti hai." Kyun sahi lagta hai: Source hi sab kuch drive karta hai. Fix: $V_{t h}$ open-circuit terminal voltage hai, jo usually internal drops ke baad source se kam hoti hai (yahan 4 V, 12 V nahi).

3. " $R_{t h}$ ke liye $R_{1}$ aur $R_{2}$ ko sirf add karo." Kyun sahi lagta hai: Woh in-line lagte hain. Fix: $V_{S}$ short karne ke baad woh terminals se parallel ho jaate hain. Sources deactivate karne ke baad hamesha redraw karo, phir series/parallel decide karo.

Recall Feynman: ek 12-saal ke bache ko explain karo

Socho ek bada vending machine hai wires se bhara hua. Tumhe sirf un do slots ki parwah hai jahan tum apne headphones plug karte ho. Andar chahe kitna bhi complicated ho, machine aise kaam karti hai jaise ek battery ek speed-bump resistor ke through push kar rahi ho. Kuch bhi plug nahi karne par voltage measure karo ( $V_{t h}$ ), aur dekho ki push karte waqt yeh kitna stiff lagta hai ( $R_{t h}$ ). Yeh do facts tumhe sab kuch bata dete hain jo machine kabhi bhi tumhare headphones ke saath karti.

Flashcards

Thevenin's theorem ek do-terminal linear network ko kisse replace karta hai?

Ek voltage source

V_{t h}

ek resistor

R_{t h}

ke series mein.

V_{t h}

kaise find karte hain?

Yeh do terminals ke across open-circuit voltage hai (load removed).

Jab koi dependent sources nahi hain toh

R_{t h}

kaise find karte hain?

Sab independent sources deactivate karo (V→short, I→open) aur terminals mein dekhte hue resistance compute karo.

Dependent sources ko

R_{t h}

ke liye deactivate kyun nahi kar sakte?

Woh circuit variables par depend karte hain; unhe kill karne se behavior badal jaata hai. Test source use karo:

R_{t h} = V_{t es t} / I_{t es t}

Thevenin source ka terminal law batao.

V = V_{t h} - I R_{t h}

R_{t h}

ke liye short-circuit relation do.

R_{t h} = V_{t h} / I_{sc}

(terminals short karo, current measure karo).

Load ko maximum power transfer ke liye condition kya hai?

R_{L} = R_{t h}

Matched load ko deliver hone wali maximum power?

P_{ma x} = V_{t h}^{2} / (4 R_{t h})

Thevenin's theorem mein linearity kyun zaroori hai?

Sirf linear networks hi woh straight-line

V

–

I

relation dete hain jo superposition ko chahiye.

Connections

Norton Equivalent Circuits — dual form: $I_{N} = I_{sc}$ , $R_{N} = R_{t h}$ ; convert via $V_{t h} = I_{N} R_{t h}$ .
Superposition Theorem — woh tool jo terminal law derive karne ke liye use hota hai.
Maximum Power Transfer Theorem — direct consequence ( $R_{L} = R_{t h}$ ).
Voltage and Current Dividers — $V_{t h}$ compute karne ke liye use hota hai.
Source Transformation — Thevenin aur Norton forms ke beech swap karo.
Nodal and Mesh Analysis — $V_{oc}$ aur $I_{sc}$ find karne ke general methods.