Five identical resistors, each of resistance R, are connected in series to a bat

Question

Five identical resistors, each of resistance R, are connected in series to a battery, and a current of 3.9 A flows in the circuit. Now, another resistor of resistance R is added to the circuit, in series with the rest of the resistors. Find the new current in the circuit. Five identical resistors, each of resistance R, are connected in parallel to a battery, and a current of 3.9 A flows in the circuit. Now, another resistor of resistance R is added to the circuit, in parallel with the rest of the resistors. Find the new current in the circuit. Five identical resistors, each of resistance R, are connected in series to a battery, and a current of 3.9 A flows in the circuit. Now, another resistor of resistance R is added to the circuit, in parallel with the battery. Find the new total current in the circuit. Suppose you know the radius of a sphere, and calculate its volume as S.02 cc. Now, you do an experiment where you fill the sphere with water and measure the volume of the water as 5.13 cc. Find the percent difference in this calculation. Suppose ten different resistors are connected in parallel. If the potential difference across the battery increases, which statement about the percent change in current is correct? The percent change in current is the same for all resistors The percent change in current is greater for the largest resistor The percent change in current is greater for the smallest resistor

Explanation / Answer

. 1.   Part 1, a: 2.   I recommend assuming a voltage and calculatingthe results based on the assumption. 3.      Assume V = 234volts 4.      5*R = (234 volts)/(3.9Ampere) =60            R= 12 5.      I = (V)/(R) = (234volts)/(6*12) = 3.25Ampere . 6.   Part 1, b: 7.   Same approach as above, assume a voltageand calculate results. 8.      Assume V = 234volts 9.      R/5 = (234 volts)/(3.9ampere) = 60ohms        R =300    10.    I = (234 volts)/(6*300) =130.0E-3 Ampere . 11.   Part 1,c: 12.   Same approach as before, except we know fromPart 1(a) that R=12. 13.   Req =(60)*(12)/(60 + 12) = 10 14.   I = (234 volts)/(10) =23.4 Ampere . 15.   Part 2, a: 16.   Percent Difference =(100%)*(new-old)/(old) 17.   Percent Difference =(100%)*(5.18-5.02)/(5.02) = 3.18725% . 18.   Part 2, b: 19.   We know from KCL and KVL that the relationshipis linear, therefore percentchange in current is the same for all resistors. .

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