2. A non-ideal battery with internal resistance r is connected as shown to a col
ID: 1520434 • Letter: 2
Question
2. A non-ideal battery with internal resistance r is connected as shown to a collection of eight identical light bulbs, each with the same resistance R bulb bulb bulb bulb bulb bulb -bulb R-3.00 bulb a. Rank the brightnesses of the bulbs, from brightest to dimmest. b. Suppose that bulb 1 could be exchanged for a new variety of bulb that is identical to the original in every respect except that its filament is 25% longer and 25% thicker (greater diameter). By what percentage would , (the voltage difference across bulb 1) change? Referring to the original circuit (as shown above): You are now offered a choice of removing either the straight wire A or the straight wire B. Which action would result in a larger change in the total current supplied by the battery (and what kind of change-an increase or a decrease)? Be sure to show all work and calculations for each alternative c.Explanation / Answer
a) Current will not flow through bulb3, bulb5,bulb6,bulb7, bulb8 as there is a zero-resistance wires A and B
Current in bulb2 and bulb4 will be equal as current of bulb1 distributed equally.
bulb1 > bulb2 = bulb4 > bulb3= bulb5=bulb6=bulb7= bulb8
b) R = pL/A, L becomes by 1.25 times and area becomes 1.25*1.25 times
so resistance R decreases by 1.25 times
Rnew = 3.00/1.25 = 2.40 ohm
Current i = 120/(0.01+2.40+1.50)
= 30.69 A
delta V1 = i*2.40
= 30.69*2.40
= 73.656 V
C) Rtotal if A is removed,
Rtotal = 0.01 + 3 + [[{3*9/(3+9)}+3]*3/([{3*9/(3+9)}+3] +3)]
= 0.01 + 3 + [2.25+3]*3/([2.25+3] +3) =4.92 ohm
If B is removed,
Rtotal = 0.01 + 3 + 1.5 + 2.25*3/(2.25+3)
= 5.796 ohm
If B is removed, more change in total current will occur,
current will decrease as Rtotal is increased from 4.51 ohm to 5.796 ohm
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