Hello Sir, thank you for your video. I have a question. When the bridge is at equilibrium we have R1 (R3 + Rab) = R2 ( Rx + Rbc) ... but how can we assume (?) at the same time that Rbc / Rab = R1/R2 as R1 and R2 are randomly chosen I guess ? (why not R2/R1 or even R3/Rx ...) ?
R1 and R2 are not randomly chosen bro, they are ratio arms whose ratio of resistance is fixed and only for such assumption the previous equation of wheat stone bridge remains true
Sir your explanation is awesome👌
Nice video ❤sir but improve the sound range coz it plays at low volume
Hello Sir, thank you for your video. I have a question. When the bridge is at equilibrium we have R1 (R3 + Rab) = R2 ( Rx + Rbc) ... but how can we assume (?) at the same time that Rbc / Rab = R1/R2 as R1 and R2 are randomly chosen I guess ? (why not R2/R1 or even R3/Rx ...) ?
R1 and R2 are not randomly chosen bro, they are ratio arms whose ratio of resistance is fixed and only for such assumption the previous equation of wheat stone bridge remains true
Thank you sir ❤
nice
Nice
Thanks