Voltage divider output, or R1 or R2 for a target voltage; the current-limiting resistor for LEDs with its nearest E12 value and power rating.
হালনাগাদ যাচাই করা উদাহরণ: ৭
R1 = 10 kΩ over R2 = 4.7 kΩ passes 31.97% of the input: 3.837 V from 12 V, drawing 816.3 µA.
A voltage divider of two resistors gives Vout = Vin × R2/(R1 + R2). The calculator solves for the output, or for R1 or R2 when you know the output you want, and shows how a load across R2 pulls the output down. An LED needs a series resistor R = (Vs − n·Vf)/I, where Vf is the forward voltage of each of n LEDs in series and I is the current you want.
The divider default, 10 kΩ over 4.7 kΩ on 12 V, gives 3.84 V and draws 0.82 mA. A red LED (Vf ≈ 2.0 V) at 20 mA on 5 V needs exactly 150 Ω, an E12 value, and the resistor dissipates 60 mW, so a ⅛ W part covers it with a 2× margin.
Exact values are rounded up to the next E12 value (IEC 60063), which keeps the LED current at or below target. Forward voltages are typical figures; datasheets vary by a few tenths of a volt, and the current is most sensitive when little voltage is left across the resistor.
যাচাইয়ের উৎস: Python 3.8 fractions: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA
যাচাইয়ের উৎস: Python 3.8 fractions: R2 = R1·Vout/(Vin − Vout) = 33000/1.7 Ω = 19.411765 kΩ
যাচাইয়ের উৎস: Python 3.8 fractions: R2‖RL = 5 kΩ, 12 × 5/15 = 4 V (6 V unloaded)
যাচাইয়ের উৎস: Python 3.8 fractions: (5 − 2.0)/0.02 = 150 Ω (in E12); P = 0.02² × 150 = 0.06 W; 2 × 0.06 ≤ 0.125
For one red LED at 20 mA, 150 Ω: (5 − 2.0 V)/0.020 A. A blue or white LED with a 3.2 V forward drop needs (5 − 3.2)/0.020 = 90 Ω, rounded up to 100 Ω in the E12 series, which gives 18 mA. Many indicator LEDs are bright enough at 5–10 mA, which doubles to quadruples the resistance.
For a single red LED at 20 mA, (12 − 2.0)/0.020 = 500 Ω, rounded up to 560 Ω (E12). That gives 17.9 mA and 0.18 W in the resistor, so use a ½ W part to keep a 2× margin. Three white LEDs in series instead (3 × 3.2 V = 9.6 V) need only 120 Ω, and the resistor wastes 48 mW.
Vout = Vin × R2/(R1 + R2), where R2 is the resistor between the output and ground. 10 kΩ over 4.7 kΩ on 12 V gives 12 × 4.7/14.7 = 3.84 V. For a target output, fix one resistor and solve for the other: R2 = R1 × Vout/(Vin − Vout), so 3.3 V from 5 V with R1 = 10 kΩ needs R2 = 19.4 kΩ.
The load sits in parallel with R2 and lowers the bottom resistance. Two 10 kΩ resistors on 12 V give 6 V unloaded, but a 10 kΩ load makes the bottom 5 kΩ and the output falls to 4 V. Keep the load at least 10 times R2 to stay within about 10%, or buffer the output; dividers suit reference and sensing inputs, not powering circuits.
Work out P = I²R and choose a standard rating at least twice that. A 150 Ω resistor passing 20 mA dissipates 0.02² × 150 = 0.06 W, so a ⅛ W (0.125 W) resistor meets the 2× margin. With one red LED on 12 V the resistor dissipates 0.18 W and needs ½ W. Common through-hole ratings are ⅛, ¼, ½, 1 and 2 W.
নির্ভুলতা আপনার ইনপুট ও পদ্ধতির অনুমানের ওপর নির্ভর করে। দশমিক গণনায় 50টি সার্থক অঙ্ক ব্যবহৃত হয়, কিন্তু আনুমানিক হিসাব, সংখ্যাগত পদ্ধতি ও উৎসের তথ্য কম নির্ভুল হতে পারে; প্রদর্শিত মান রাউন্ড করলে এই সীমাবদ্ধতাগুলি দূর হয় না। স্বতন্ত্র উৎসের সমাধানের সঙ্গে যাচাই করা উদাহরণ: ৭। যেমন, “12 V with 10 kΩ over 4.7 kΩ” উদাহরণটি Python 3.8 fractions: 12 × 4.7/14.7 = 3.8367347; I = 12/14700 A = 0.8163265 mA-এর সঙ্গে যাচাই করা হয়।
Horowitz & Hill, The Art of Electronics (3rd ed.), §1.2.3 Voltage dividers and §2.1 (LED current limiting); IEC 60063:2015 — Preferred number series (E12); OpenStax University Physics Volume 2, §10.2 Resistors in series and parallel.
এই ক্যালকুলেটরে স্বাধীন উৎস থেকে উত্তর নেওয়া ৭টি সমাধান করা উদাহরণ রয়েছে। এগুলি পরীক্ষার সমষ্টিতে চালানো হয়, আর আপনি এখানেও চালাতে পারেন।
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