How many 176 resistors are required
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. WebTo calculate the number of 176 ohm resistors required in parallel to achieve a specific resistance, you need to use the following formula: N = R_desired / R_single Where N is the …
How many 176 resistors are required
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WebFind 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 ... 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.
WebJun 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 WebHow 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 …
WebApr 29, 2024 · Q 10: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? From Ohm’s law, ∴four resistors of 176 Ω are required to draw the given amount of current. When three equal resistance are connected in parallel the effective resistance is 1 by 3 home if all are connected in series the effective resistance is? 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 …
WebJun 24, 2024 · According to Ohm’s law, V = IR Now for x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω. Therefore, …
WebMar 21, 2024 · Note that you can connect two resistors ( R1 and R2) in two ways: parallel and series. The resulting resistance ( Rp and Rs respectively) in those cases are: Rs = R1 … churches midland miWebMar 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 … de vere golf club blackpoolWebMar 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 de vere horsley estate east horsleyWebI ( 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 / … de vere horsley estate ockham road southWebSolution 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 de vere hotel portsmouthWebApr 8, 2024 · Complete step by step answer: We have given here to find the number of \ [176\Omega \] resistors needed to carry \ [5A\] on a \ [220V\] line. Let, the number of resistors needed is \ [x\]. Hence, the equivalent resistance of the \ [x\] number of resistors will be when connected in parallel, churches midlothian txWebMar 21, 2024 · The continued fraction (="Euclidean"?) gives contfrac(x) = [0, 1, 5] so we have x = 1 / (1 + 1 / 5) and thus x = r 5r and we need 6 resistors, which is not optimal. A better configuration is by 5 / 6 = 1 / 2 + 1 / 3 = 2r ⊕ 3r so we need 5 resistors. de vere hotel loch lomond scotland