could use some help, thanks. CYBERPOWERPO HY 221 RESISTORS IN SERIES AND IN PARA

Question

could use some help, thanks.

CYBERPOWERPO HY 221 RESISTORS IN SERIES AND IN PARALLEL Name ive: In this laboratory, we investigate the rules for combining resistors in series and in parallel DC power supply Ammeter and voltmeter (or 2 digital multimeters) Three (3) different linear resistors quipment: Connecting wires and alligator clips Background A. Resistors connected in series (see Fig. 1) can be treated as though the network effective resistance Rs is the sum of the individual resistances were a single resistor whose R: Resistors connected in parallel (see Fig. 2) can be treated as though the network were a single resistor whose effective resistance Re is the given by: (Eq. 2) Fig. I Fig. 2 For each network shown in the figures above, suppose that resistor R, had the least resistance and R had the greatest resistance fi.e, R,R,RI. 1. Do you think that the equivalent resistance Rs of the series network (shown in Fig. 1) would be less than Ri, greater than R, or between R, and R,? Explain your reasoning. Do you think that the equivalent resistance Rp of the parallel network (shown in Fig. 2) would be less than R, greater than R, or between R and R,? Explain your reasoning. If the series and parallel networks shown above were connected to identical batteries, do you think the total current through the series network would be greater than, less than, or equal to the total current through the parallel network? Explain your reasoning. 2. [Note: In the summary questions for this lab, you will be asked to review and, if necessary, revise your answers on this page. If you are unsure of any answers, that is okay, keep working!]

Explanation / Answer

1. in series effective resistance is given by R = R1 +R2 + R3

effective resistance would be greater than any individual resistance i.e. R1, R2, R3 (because it is sum of all resistance)

in parallel combination effective resistance is 1/R = 1/R1 + 1/R2 + 1/R3

effective resistance will be less than the smallest resistance.

i.e. R < R1 , R < R2 , R < R3 (because inverse of effective resistance is sum of inverse of individual resistance).

2.

if a battery of voltage V is connected across series combination then current is given by

i = V/R = V/(R1+R2+R3)

while current in parallel combination is given by

i = V*(R1R2+R2R3+R3R1)/(R1R2R3)

current in parallel combination will be more than series combination because effective resistance in series combination is more and current is inversely proportional to effective resistance.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.