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DC circuits
1. Battery Model
2. Simple series and parallel
3. Combo series and parallel
4. RC circuits
5. Kirchhoff’s Laws
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Animation - DC Circuits
http://faraday.physics.utoronto.ca/IYearLab/Int
ros/DCI/Flash/WaterAnalogy.swf
• Pressure drops (voltage drops) all add
• Flow rate in = flow rate out (current in = current out)
Battery Model
• Voltage source and internal resistance
–
–
–
–
Source electrochemical “emf”
Internal resistance
Resistance increases with time (battery goes bad)
Model is series of voltage and resistance
Example 19-1 (1)
• Find current (voltages add to 0)
12 𝑉 − 𝑉𝑟 − 𝑉𝑅 = 0
12 𝑉 − 𝐼 0.5 Ω − 𝐼 65Ω = 0
𝐼=𝑅
𝑉
𝑡𝑜𝑡𝑎𝑙
•
12 𝑣
= 65.5Ω = 0.183 𝐴
Find voltage drops
𝑉𝑟 = 𝐼 0.5 Ω = 0.0915 V
𝑉𝑅 = 𝐼 65 Ω = 11.895 V
11.895 V + .0915 V = 12 V
Example 19-1 (2)
• Find voltage across battery
𝑉 = ℰ − 𝐼𝑅ℰ = 12𝑉 − (0.183 𝐴) (0.5 Ω)
• Find Power consumed by resistors
•
𝑃𝑟 = 0.183 𝐴
2
0.5 Ω = 0.02 𝑊
𝑃𝑅 = 0.183 𝐴
2
65 Ω = 2.18 𝑊
Find Power delivered by battery
𝑃𝑏𝑎𝑡 = 𝐼𝑉 = 0.183 𝐴
12 𝑉 = 2.20 𝑊
Simple Series Circuit
• Rules for Series
– Voltages add
– Current same
• Equivalent circuit
– 𝑉𝑡𝑜𝑡 = 𝑉1 + 𝑉2 + 𝑉3 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 = 𝐼 𝑅1 + 𝑅2 + 𝑅3 = 𝐼𝑅𝑒𝑞
– Equivalent series resistance
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3
– Example (12 v battery, 4 Ω resistors)
Simple Parallel Circuit
• Rules for Parallel
– Voltage same
– Currents add
• Equivalent circuit
– 𝐼𝑡𝑜𝑡 = 𝐼1 + 𝐼2 + 𝐼3 =
𝑉
𝑅1
+
𝑉
𝑅2
+
𝑉
𝑅3
– Equivalent parallel resistance
=𝑉
1
𝑅1
1
𝑅𝑒𝑞
– Example (12 v battery, 4 Ω resistors)
+
=
1
𝑅2
1
𝑅1
+
+
1
𝑅3
1
𝑅2
=
+
𝑉
𝑅𝑒𝑞
1
𝑅3
Series vs. Parallel
• Criterion for series and parallel (do not confuse!)
• Weird geometries
Simple Series and Parallel
• Example 19-3 Parallel, R = 100 Ω
– Find I through each resistor (V same)
– Find total current I = I1 + I2 ,
– Find apparent R = V/I
1
𝑅𝑒𝑞
=
1
𝑅1
+
1
𝑅2
• Example 19-3 Series R = 100 Ω
– Add V across both resistors (I same)
– Find common current 𝐼𝑅1 + 𝐼𝑅2 = 𝑉
– Find apparent R = V/I
𝑅𝑒𝑞 = 𝑅1 + 𝑅2
Combo Series and Parallel - 19.4
• Progressively simplify
– Find equivalent resistances
– Parallel, series, series
– Total resistance 690 Ω
• Find current through battery
12 𝑉
𝐼 = 690 Ω = 17 𝑚𝐴
• Reconstruct voltage drops currents
– Current through 400 Ω (same as battery)
– Voltage across 400 Ω
– Voltage across 500/700
– Currents through 500/700
– etc, etc.
Combo Series and Parallel 19.5
• Build it back up
– Find voltage drop near battery
– Find other voltage drops
– Find currents, etc, and continue
• Current through battery 17 mA
• Current through 400 Ω also 17 mA
• Voltage across 400 Ω
17 𝑚𝐴 400 Ω = 6.8𝑉
• Voltage across 500/700
12 V – 6.8 V = 5.2 V
• Currents through 500/700
I1=10 mA
I2=7 mA
Combo Series and Parallel 19.7
• Simplify down
– Rtot = 10.3 Ω
– Ibat = 0.87 A
• Build back up
– Voltage across top
= 9 𝑉 − .87 𝐴 .5 Ω −
.87 𝐴 5Ω
– Current through 10 Ω
– Current through 8.7 Ω
– Current through 6 Ω
– Voltage across 4/8 Ω
More complicated series parallel
• Example 19-7
– Current from battery
– Drop across battery
– Current in 6Ω resistor
• Solution
–
–
–
–
Get total resistance (break down)
Get total I
Establish IR drops (build up)
Find current
• Like solving crossword puzzle!
Car Talk Puzzler*
Go out to start your car. New battery, radio works, but electrical goes dead
when you try to start.
• Starter Motor
– Estimate power (1600 W)
– Estimate current (133 A)
– Estimate resistance (.09 ohms)
• Radio
– Estimate power (100 W)
– Estimate current (8 A)
– Estimate resistance (1.5 ohms)
• Draw circuit, parallel starter/radio in series with battery and cable/contact
resistance. (0.1 ohms)
*Apologies to Click & Clack
Example - Problem 20
More examples
• Chapter 19 - 13, 14, 17, 19, 20 - Resistors in
series and parallel