Calculus Parallel Resistors Question Help?
Need some detailed help with the following problem please!
An electrical circuit consists of two parallel resistors, with resistances R1 and R2 respectively. The total resistance R of the circuit (measured in Ohms) is specified by (1/R = (1/R1) + (1/R2).
The resistors are heating up, so their resistances are increasing over time. Suppose that R1 is increasing at a rate of .3 Ohms/s and R2 is increasing at a rate of .2 ohms/s.
When R1 = 80 Ohms and R2 = 100 Ohms, how fast is the total resistance increasing?
- Anonymous8 years agoFavorite Answer
I have posted a full solution to your question at one of the math help forums I help to moderate so that I may provide you with an easy to read explanation using LaTeX:
- fesmireLv 44 years ago
once you have 2 factors (batteries, resistors, something) in parallel. The voltage for the time of them will continuously be the comparable. so that it is going to discover the flexibility draw of two resistors you will use voltage for the time of them, expanded by the finished contemporary (I) by using the two one among them. So 2 resistors in parallel each and each drawing 2 amps, with 10 volts for the time of them. would draw (2+2)A * 10V = 40 Watts. it may help to comprehend in case you broke up the priority, and solved each and all of the resistors in my view, and then further to get finished ability utilization. So for resistor a million; 2Amps @ 10V = 20 Watts. For resistor 2; 2 Amps @ 10V = 20 Watts. finished ability intake = 20W + 20W = 40 Watts. it may additionally help to think of of the resistors as a "black container". considering the fact that your ability equation P=IV will continuously be surprising. you are able to basically seem on the voltage enter on your regular equipment, and the present enter into your regular equipment, and resolve for finished ability intake. So, 4 Amps in situations 10V in, equals 40 Watts.