How many 176 resistors are required

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August undefined, 2024

WebApr 30, 2024 · 4 Resistors are required. Step-by-step explanation: Given: Volts = 220 Volts. Current = 5 Ampere. Resistance of each resistor, R = 176 Ohms. No of resistors = n. We know that: v = RpI. ∴ ∵ . Taking N to the other side we … Web1 R = x× 1 (176) 1 R = x × 1 ( 176) R = 176 x 176 x. Now, as per Ohm’s Law, we get, V I = 176 x V I = 176 x. ⇒ x = 176×5 220 = 4 x = 176 × 5 220 = 4. Hence, in order to draw the required amount of current, four 176 Ω resistors will be needed. Related Questions:

How many $176\\Omega $ resistors (in parallel) are required

WebOct 3, 2024 · We know that according to Ohm’s Law, V = IR. Therefore, R = V/I = 220/5 = 44ohm. Now since n such resistors are connected in parallel,so. 1/R = n/176. or, 1/44 = … WebAnd then we get One upon. Our total is going to want upon 44 because the maximum resistance will be 44. So and upon 1 76. If we solve this, If we solve this then we get and … hill country power dripping springs tx

How many $176\\Omega $resistors (in parallel) are required to …

WebApr 26, 2024 · Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?Welcome to our channel, hope you will get sufficient information from o... WebMar 29, 2024 · (a) For highest resistance according to question resistances must be connected in series: 4 Ω, 8 Ω, 12 Ω, 24 Ω R 1 = 4 Ω R 2 = 8 Ω R 3 = 12 Ω R 4 = 24 Ω Total resistance in series = R 1 + R 2 + R 3 + R 4 = 4 + 8 + 12 + 24 = 48 Ω Now, resistance is maximum when connected in series. WebMar 9, 2024 · Now for x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω. 1/44 = 1/176 + 1/176 + 1/176 .....χ times. 1/44 = x / 176 => x = 176/ 44 => x = 4. Therefore, 4 resistors of 176 Ω are required to draw the given amount of current. I hope, this will help you___ . Thank you hill country premier rentals

How many 176ohm resistors(in parallel) are required to carry

Answered: How many 176Ω resistors (in parallel)… bartleby

WebClick here👆to get an answer to your question ️ How many 176 Ω ?resistor(in parallel) are required to carry 5 A in 220 V line. Solve Study Textbooks Guides. Join / Login. Question . ... Calculate the equivalent resistance of the two resistors of 100 ... WebApr 8, 2024 · Hint: The equation for the net resistance for two resistors of resistances connected in parallel hast to be used first. Further, Ohm's law and Kirchoff’s current law will also be used to find the number of resistors. Formula used: The net resistance for two resistors of three resistances \[({R_1}),({R_2})\]and \[\left( {{R_3}} \right)\] connected in … smart array とはWebHow Many 176 ω Resistors (In Parallel) Are Required to Carry 5 a on a 220 V Line? CBSE English Medium Class 10. Question Papers 939. Textbook ... Therefore, four resistors of 176 Ω are required to draw the given amount of current. Concept: System of Resistors - Resistances in Parallel. hill country properties brady tx

"WebApr 5, 2024 · Therefore, current through 12 Ω resistor = 0.67 A. 12. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? Suppose n resistors of 176 Ω are connected in parallel. 1/R = 1/176 + 1/176 + ……….n times 1/R = n/176 or R= 176/n Ω By Ohm’s law R=V/I 176/n = 220/5 n= (176 x 5)/220 = 4 13. " - How many 176 resistors are required

How many 176 resistors are required

How do you find the resistance? – ProfoundTips

WebQ 10: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? Ans: For x number of resistors of resistance 176 Ω, Here Supply voltage, V = 220 V Current, I = 5 A Equivalent resistance of the combination = R, given as From Ohm’s law, ∴four resistors of 176 Ω are required to draw the given amount of current. Important Links WebMar 14, 2024 · Hence using the Ohm’s law and the value of resistance from equation (1), we get: V = I R n e t ⇒ 220 = 5 ( 176 N) ⇒ N = ( 176 44) ⇒ N = 4. Hence, 4 of the 176 Ω resistors in parallel would be required to get a net current of 5A flowing through a 220V line.

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WebNow, as per Ohm’s Law, we get, V I = 176 x V I = 176 x ⇒ x = 176×5 220 = 4 x = 176 × 5 220 = 4 Hence, in order to draw the required amount of current, four 176 Ω resistors will be … WebAre you wondering how many 176 ohm resistors in parallel are required for your circuit? This is a common … Read more. Resistor color codes. are a standard system for indicating the value of resistors in electronic circuits. Understanding how to read these codes is an essential skill for anyone working with electronic components, whether you ...

WebMar 19, 2024 · For three resistors if we connect them in parallel the equivalent resistance is R 1 R = 1 R1 + 1 R2 + 1 R3 1 R = 1 R 1 + 1 R 2 + 1 R 3 For x number of resistors of resistance 176 Ω the equivalent resistance of the resistors connected in parallel is given by ohm’s law as V = IR R = V I V I Where,supply voltage, V = 220 V current I = 5A WebSolution for How many 176Ω resistors (in parallel) are required to carry 5 A on a 220 V line? Skip to main content. close. Start your trial now! First week only $4.99! arrow ... How many 176Ω resistors (in parallel) are required to carry 5 A on a 220 V line? Expert Solution. Want to see the full answer? Check out a sample Q&A here.

WebQuestion From - NCERT Physics Class 10 Chapter 12 Question – 033 ELECTRICITY CBSE, RBSE, UP, MP, BIHAR BOARDQUESTION TEXT:-How many `176 Omega` resistors (in... WebAns: For x number of resistors of resistance 176 Ω, Here Supply voltage, V = 220 V Current, I = 5 A Equivalent resistance of the combination = R, given as From Ohm’s law, ∴four …

WebR_total = (R1 * R2) / (R1 + R2) Where R1 and R2 are the values of the two resistors. To calculate the number of 176 ohm resistors required in parallel to achieve a specific resistance, you need to use the following formula:

WebMar 24, 2024 · Best answer Let the resistors be x’ Resistance = 176Ω As per ohm’s law . V = IR R = V I V I Here, V = 220 V, I = 5A ∴ Four resistors (in parallel) are required to carry 5A on a 220 V line. ← Prev Question Next Question → Find MCQs & Mock Test JEE Main 2024 Test Series NEET Test Series Class 12 Chapterwise MCQ Test Class 11 Chapterwise Practice … hill country pump service fredericksburg txWebSolution Verified by Toppr Step 1 : Equivalent resistance Let n be the number of resistors to be connected in parallel Then, R 1=R 2=......=R n=R=176 Ω Now, R eq1 = R 11+ R 21 +..... R n1 ⇒ R eq1 = R 1n= 176n ⇒R eq=176/n ....(1) Step 2 : Applying Ohm’s law V=IR eq ⇒220=5R eq ⇒220=5×176/n ( Using Equation (1)) ⇒n=4 smart arse imagesWebHow many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? DARSHAN CLASSES 9,159 views Apr 26, 2024 Q10. How many 176 Ω resistors (in parallel) are … smart arrowWebMar 30, 2024 · NCERT Question 10 How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? Current = I = 5 A Potential difference = V = 220 V Resistance of … hill country property for sale by ownerWebJun 17, 2016 · Now for x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω. 1/44 = 1/176 + 1/176 + 1/176 .....χ times. 1/44 = x / 176 => x = 176/ 44 => x = 4. Therefore, 4 resistors of 176 Ω are required to draw the given amount of current. I hope, this will help you___ . Thank you hill country pump service fredericksburgWebI ( current ) = V (voltage ) / R ( resistance ) . hence to carry 5A at 220 V supply Resistance required V / I = 220 / 5 = 44 OHMS. In parallel circuit 1/R = 1/R1 + 1/ R2 + 1 / R3 so on …. Hence 1 / 44 = n / 176 when n = no of parallel resistance so n = 176 / … smart arse synonymWebFind many great new & used options and get the best deals for Ohmite Two Ganged 35 Ohm 1.0 Amp Wire Wound Potentiometers at the best online prices at eBay! ... Vitreous Enamel Adjustable Wire-Wound Power Resistors, 100 Watts. $5.59 + $4.50 shipping. Ohmite 12 Ohm 2.04 Amp Wire Wound Potentiometer ... Minimum monthly payments are required ... smart art access