Peoples Physics Concepts

http://www.ck12.org Chapter 13. Electric Circuits: Batteries and Resistors

Question:What is the total resistance of the circuit?

Answer:In order to find the total resistance we do it in steps (see pictures. First add the 90Ωand 10Ωin series to

make one equivalent resistance of 100Ω(see diagram at below). Then add the 100Ωto the 10Ωin parallel to get

one resistor of 9. 1 Ω. Now we have two resistors in series, simply add them to get the total resistance of 29. 1 Ω.

Question:What is the total current coming out of the power supply?

Answer:Use Ohm’s Law(V=IR)but solve for current(I=V/R).Itotal=VRtotaltotal= 20 V/ 2. 91 Ω= 0. 69 Am ps

Question:What is the power dissipated by the power supply?

Answer:P=IV, so the total power equals the total voltage multiplied by the total current. Thus,

Ptotal=ItotalVtotal= ( 0. 69 A)( 20 V) = 13. 8 W.

Question:How much power is the 20Ωresistor dissipating?

Answer:The 20Ωhas the full 0.69Amps running through it because it is part of the ’main river’ (this is not the

case for the other resistors because the current splits).P 20 Ω=I 202 ΩR 20 Ω= ( 0. 69 A)^2 ( 20 Ω) = 9. 5 W

Question:If these resistors are light bulbs, order them from brightest to least bright.

Answer:The brightness of a light bulb is directly given by the power dissipated. So we could go through each

resistor as we did the 20Ωguy and calculate the power then simply order them. But, we can also think it out. For

the guys in parallel the current splits with most of the current going through the 10Ωpath (less resistance) and less

going through the 90Ω+ 10 Ωpath. Well the second path is ten times the resistance of the first, so it will have one

tenth of the total current. Thus, there is approximately and 0.069 Amps going through the 90Ωand 10Ωpath and

0.621Amps going through the 10Ωpath.

P 10 Ω=I 102 ΩR 10 Ω= ( 0. 621 A)^2 ( 10 Ω) = 3. 8 W

P 90 + 10 Ω=I 902 + 10 ΩR 90 + 10 Ω= ( 0. 069 A)^2 ( 100 Ω) = 0. 5 W