1.2.10 · Hardware › Circuit Analysis Fundamentals
Koi bhi uljha hua linear two-terminal network — chahe kitne bhi resistors aur sources hon — apne do output terminals ke nazariye se single current source in parallel with a single resistor ki tarah behave karta hai. Woh pair hi Norton equivalent hai. Mess ko replace karo, behavior ko rakho.
Intuition Yeh sach kyun hona hi chahiye
Ek linear network ka terminal behavior terminal voltage V aur terminal current I ke beech ke relationship se puri tarah describe hota hai. Ek linear circuit ke liye yeh relationship ek straight line hoti hai: I = a V + b . Ek straight line ko sirf do numbers chahiye (slope aur intercept). Toh koi bhi linear network us line ko reproduce karne ke liye sirf do components lega:
ek current source (intercept → terminals short hone par current),
ek resistor (slope → jaise voltage badhti hai current girti hai).
Yahi exactly ek current source in parallel with a resistor hai. Describe karne ke liye kuch bhi nahi bachta.
Definition Norton equivalent
Ek linear two-terminal network ek current source I N in parallel with a resistance R N ke equivalent hai, jahan
I N = Norton current = woh current jo terminals ke across rakhe gaye short circuit se flow karti hai.
R N = Norton resistance = terminals mein jhank kar dikhai dene wali resistance jab saare independent sources deactivate kar diye jaayein (voltage sources → short, current sources → open).
Worked example 3-step method
Step 1 — I N nikalein: do output terminals ko short karo; us short se guzarne wali current compute karo.
Yeh step kyun? Short V = 0 force karta hai, toh I L = I N − 0 = I N . Short seedha intercept padh leta hai.
Step 2 — R N nikalein: saare independent sources band karo (V-source → wire, I-source → gap) aur terminals ke beech equivalent resistance nikalo.
Yeh step kyun? I –V line ki slope sirf resistors set karte hain; sources sirf line shift karte hain, use tilt nahi karte. Unhe deactivate karna pure resistance ko isolate karta hai.
Step 3 — Draw karo I N ∥ R N . Yahi tumhara equivalent hai.
Worked example Do resistors aur ek source
Ek 12 V source R 1 = 4 Ω ko series mein feed karta hai ek node tak; us node se R 2 = 6 Ω bottom rail tak jaata hai. Output terminals R 2 ke across hain. Norton equivalent nikalo.
Step 1 (I N ): Output terminals ko short karo. Yeh R 2 ko short kar deta hai (uske dono ends ab jud gaye), toh saari current short se jaati hai:
I N = 4 12 = 3 A .
Kyun? Short ke saath, R 2 par koi voltage nahi, toh woh invisible hai; sirf R 1 source ko limit karta hai.
Step 2 (R N ): Source ko kill karo (12 V → wire). Ab R 1 aur R 2 dono terminals ko connect karte hain — woh parallel mein hain:
R N = 4 + 6 4 ⋅ 6 = 2.4 Ω.
Parallel kyun? Source deactivate karne se R 1 aur R 2 ke tops aapas mein jud jaate hain aur terminals R 2 ke across hain, toh dono resistors same do nodes bridge karte hain.
Result: I N = 3 A ∥ R N = 2.4 Ω .
Worked example Ek load lagao aur forecast karo
Upar wale equivalent ke saath R L = 2.4 Ω load attach karo.
Current R N (2.4 Ω) aur R L (2.4 Ω) ke beech barabar split hogi:
I L = I N ⋅ R N + R L R N = 3 ⋅ 4.8 2.4 = 1.5 A .
Current-divider kyun? Do equal resistances 3 A barabar share karti hain. Forecast: equal resistors ⇒ aadha aadha ⇒ 1.5 A. ✓ Verified.
Load voltage: V = I L R L = 1.5 × 2.4 = 3.6 V .
R N nikalte waqt sources ON rakhna
Kyun sahi lagta hai: "Sources circuit ka hissa hain, toh unhe rakho." Kyun galat hai: R N ek slope hai, purely resistive; live sources ek fixed offset add karte hain jo simple resistance reading corrupt kar deta hai. Fix: independent sources deactivate karo (V→short, I→open) R N compute karne se pehle. (Dependent sources rakho — unke liye test source method use karo.)
I N ke liye R 2 short ho jaata hai, yeh bhool jaana
Kyun sahi lagta hai: R 2 ek real resistor hai, zaroor current limit karega. Kyun galat hai: terminals ke across short R 2 ke parallel mein 0 Ω path rakhta hai, saari current uske around divert kar deta hai. Fix: short ke saath redraw karo; jo bhi element short ke parallel hai use delete kiya ja sakta hai.
I N ya R N compute karte waqt load resistor use karna
Fix: Norton equivalent woh network hai jo load minus hai. Pehle load hataao, phir I N aur R N nikalo; tabhi load reconnect karo.
Recall Feynman: ek 12-saal ke bachche ko samjhao
Socho ek water tank pipes se tumhare sink ko feed karta hai. Har pipe draw karne ki jagah, main tumhe ek card deta hoon: "yeh tap 3 cups per second dhakelta hai, aur ek leaky pipe hai 'thickness 2.4' ki jo kuch chura leti hai." Ab tum predict kar sakte ho kitna paani kisi bhi cup tak pahunche ga — tank dobara dekhe bina. Norton = circuit ke outlet ka sabse chhota imeandaar summary.
"Current ke liye short, resistance ke liye kill."
I N : terminals short karo. R N : sources kill (deactivate) karo. Phir N orton draw karo = current source N ext to (parallel) ek resistor.
Ek linear two-terminal network ka Norton equivalent kya hota hai? Ek current source I N in parallel with a resistance R N .
Norton current I N kaise nikalte hain? Output terminals short karo aur short se guzarne wali current compute karo.
Norton resistance R N kaise nikalte hain? Saare independent sources deactivate karo (V→short, I→open) aur terminals ke beech resistance nikalo.
Norton terminal equation kya hai? I L = I N − V / R N .
R N aur R T h mein kya relation hai?Dono equal hain: R N = R T h .
Thévenin ko Norton mein convert karo. I N = V T h / R T h , aur R N = R T h .
Linear network ke liye terminal I–V relationship straight line kyun hoti hai? Superposition/linearity I ko V ka affine function banati hai: I = aV + b , jo sirf do parameters maangti hai.
I N nikalte waqt terminals ke across resistor ko ignore kyun kiya ja sakta hai?Parallel mein short saari current le jaata hai, toh us par 0 V hota hai aur woh invisible ho jaata hai.
Norton source se load R L mein current-divider? I L = I N ⋅ R N / ( R N + R L ) .
Thévenin equivalent circuits — voltage-source twin; freely convert karo.
Source transformation — bridge V T h = I N R N .
Ohm's Law — V / R N leak term provide karta hai.
Kirchhoff's Current Law — node par I L = I N − V / R N justify karta hai.
Maximum power transfer — max load power ke liye R L = R N use karta hai.
Superposition — kyun terminal behavior linear/affine hai.
Linear two-terminal network
Terminal law I equals aV plus b straight line
I_L equals I_N minus V over R_N