1a) Two light bulbs are hooked up in a series to a 9 V battery. Why is the voltage drop across a single light bulb not 9 V? 1b) Suppose someone hands you three resistors having three different resistances. How would you hook them up in order to make their equivalent resistance as small as possible, and thus the largest current? (Briefly explain why your arrangement produces the smallest equivalent resistance.) Solution In the series circuit, any current that flows through one bulb must go through the other bulbs as well, so each bulb draws the same current. Since if both bulbs are light bulbs, they have the same resistance, so the voltage drop across each one is the same and equals but not equal to 9 v v=ir ; there will be some resistance in bulb which causes to change voltage drop 2. I will hook them in parallel series. because of the equivalent resistance are small in this series 1/R eq = 1/R1 + 1/R2 ........ the current may not be the same through all the branches in the parallel network. However, the voltage drop across all of the resistors in a parallel resistive network is the same. I= V/R EQ Req is smaller. current will be larger. .

1. 1a) Two light bulbs are hooked up in a series to a 9 V battery. Why is the voltage drop across a
single light bulb not 9 V?
1b) Suppose someone hands you three resistors having three different resistances. How would
you hook them up in order to make their equivalent resistance as small as possible, and thus the
largest current? (Briefly explain why your arrangement produces the smallest equivalent
resistance.)
Solution
In the series circuit, any current that flows through one bulb must go through the other bulbs as
well, so each bulb draws the same current. Since if both bulbs are light bulbs, they have the same
resistance, so the voltage drop across each one is the same and equals but not equal to 9 v
v=ir ; there will be some resistance in bulb which causes to change voltage drop
2. I will hook them in parallel series. because of the equivalent resistance are small in this series
1/R eq = 1/R1 + 1/R2 ........
the current may not be the same through all the branches in the parallel network. However, the
voltage drop across all of the resistors in a parallel resistive network is the same.
I= V/R EQ Req is smaller. current will be larger.