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10.5 CYU Suggested Answers pg. 325
1. The total voltage of resistors in series is equal to the sum of the individual voltages. The total voltage
of resistors in parallel is equal to the voltage of any one of the resistors (the resistors all have the same
voltage).
2. (a) Answers may vary. A sample circuit diagram is shown (the battery is made of six cells that each
have a voltage of 1.5 V).
(b) Since the resistors are in series, they each get 2.25 V (or one quarter of the 9 V). Using this and
Ohm’s law gives 0.10 A in each resistor.
(c) The total resistance is 22 Ω x 4 = 88 Ω.
3. (a) The voltage of each resistor is 120 V.
(b) The current in each resistor is 0.6 A.
(c) The resistance of each resistor is 200 Ω.
(d) The total resistance is 100 Ω.
4. (a) The current in the second light bulb is 280 mA.
(b) The light bulbs are not identical. Since they have equal voltages and different currents, according to
Ohm’s law they have different resistances.
(c) The resistance of the first light bulb is R1 = 19 Ω; the second light bulb is R2 = 11 Ω; and the total
resistance is RT = 6.8 Ω.
(d) There will be no voltage (or V2 = 0 V). The opened switch disconnects the light from the circuit—
there is no complete path.
(e) The voltage across the first bulb is still 3 V
(f) The current delivered by the battery is IT = 160 mA.
5. The total resistance of the three resistors in series is 180 Ω.
6. The diagram shows the placement of the meters.
7. The diagram shows the placement of the meters.
8. (a) The voltages of the resistors are V1 = 6 V, V2 = 6 V, and V3 12 V.
(b) The current in a resistor is inversely proportional to the resistance. Since the resistance of the first
branch is twice that of the second branch, the current will be half, or 250 mA.
(c) The total current delivered by the battery is 250 mA + 500 mA = 750 mA. Therefore, the total
resistance is 16 Ω.