1. Two identical resistors are wired in series. An electrical current runs through the combination. If the current through the first resistor is I 1, then the current through the second is? 1) also I 1 2) I 1 3) smaller than I 1, but not necessarily I 1

2. As more identical resistors R are added to the parallel circuit shown, the total resistance between points P and Q 1) increases. 2) stays the same. 3) decreases.

3. The circuit below consists of identical light bulbs burning with equal brightness, and a single 12 V battery. When the switch is closed, the brightness of bulb A 1) increases. 2) stays the same. 3) decreases.

4. The four bulbs are identical. Which circuit puts out more total light? 1) I. 2)Both circuits emit the same amount of light. same amount of light. 3) II.

5. The three bulbs in the circuit all have the same resistance. Their brightness indicates the power being dissipated by each. 1. B shines brighter than C. 2. B and C are equally bright, each twice as bright as A. each twice as bright as A. 3. B and C are equally bright, each half as bright as A. each half as bright as A. 4. B and C are equally bright, each ¼ as bright as A. each ¼ as bright as A. 5. All three bulbs burn with equal intensity. equal intensity.

6. - + 33 66 22 12v Total effective resistance of this circuit? Total current drawn? At what rate is the (entire) circuit dissipating energy?

7. 1)also I 1 3)decreases. 1)Increases 3)II. 4. B and C are equally bright, each ¼ as bright as A. each ¼ as bright as A. Answers to slides The bottom branch draws half as much current I bottom =V/(2R) compared to I top =V/R. Lets call I bottom = I and then I top = 2 I. Bulb A burns with P=(2 I ) 2 R while B or C burns Individually with P= I 2 R. The bottom branch as a whole has two bulbs burning 2 I 2 R, but we are asked to compare each bulb. one more slide

8. - + 33 66 22 12v Total effective resistance of this circuit? Total current drawn? At what rate is the (entire) circuit dissipating energy? 12v/4  = 3 A 1/R pair = 1/3  + 1/6  R total = R pair +2  = 4  P=IV=(3 A)(12v) = 36 watts