Critical Thinking

3 – 14 - 2013

#16 Answer• 16a-The voltage of the two batteries in a series is found by summing

the voltage of each battery. Vt=V1 + V2 or 1.5V + 1.5V = 3V The voltage supplied by the two batteries in parallel is 1.5V and this is the voltage dropped across the bulb.• 16b-The current through the buld with the batteries in a series

recieves twice the current as that of the bulb with the batteries in parallel, since the voltage is twice as much. The resistance stays the same, so if the voltage doubles the current doubles.

• 16c- Use Ohm’s Law-I=V/R or 3V/12 Ω = 0.25A for the circuit with the batteries in a series.

• OR• Again use Ohm’s Law for the Parallel circuit but this time 1.5V/12 Ω

=0.125A

#17 Answer

• 17a-The current flowing through the blender is I=P/V or 300W/120V=2.5A

• 17b-The resistance of the frying pan can be found from R=V2/P or 120V2/1200W=12 Ω

• With a circuit breaker of 20A, the maximum power drawn must be less than or equal to Pmax=Vmax or 120V*20A=2400W

• So your options are– Toaster/frying pan (2200W)– Toaster/coffee maker (1900W)– Frying pan/blender/coffee maker (2100W)

#18 Answer

• A. Use Q=mcΔT=• M=500g• Tinitial=60 °C• Tfinal = 50 °C• C=4180 J/kg°c• 0.500kg*4180* (50-60)=-20,900J• B. In a closed system the cold water should gain all the

heat energy transferred away from the hot water or 20,900J. This is assuming that no energy was transferred to the surroundings.

18 C

• Tcold=10 °C• Thot=60 °C• Tfinal=40 °C• Mhot=500g or 0.5kg• So using mcΔT we can say that• MhotcΔT + McoldcΔT = 0• Or• 0.5kg*4180*(40 °C -60 °C)+Mcold*4180*(40 °C -10 °C)=0• 0.5*-20 °C=-Mcold*(-30 °C)• Mcold=0.33kg

Answer for #20

• A. The voltage across the light bulb is the same as the voltage of the battery or 15V

• B. The Current through the resistor can be found by using Ohm’s Law. I =V/R– Or, 15V/6Ω=2.5A

• C. Find total current and then apply Ohm’s law– I=V/R or 15/9=1.66 and 15/6=2.5 – Total current = 4.1– Then find resistance which is 15/4.1=3.6 Ω

• D. When the switch is closed the bulb will go out because the electricity will flow through the 0 Ω resistor

Answer to #21

Answers to #23

• A=1.7 Ω• B=1.3A• C=21W