Physics Unit 9

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Charge in 1 hour:𝐼 = Δ𝑄/Δ𝑡 Δ𝑄 = 𝐼Δ𝑡 = .2 𝐴 3600 𝑠 = 720 𝐶

Energy:𝐸𝑃𝐸 = 𝑞𝑉 = 720 𝐶 12 𝑉 = 8640 𝐽

The speakers usually don’t draw that much current. They only draw that much current at their maximum volume.

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𝑉 = 𝐼𝑅 → 12 𝑉 = 0.20 𝐴 𝑅 → 60 Ω = 𝑅

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𝑅 =𝜌𝐿

𝐴

𝑅

𝐿=𝜌

𝐴𝜌 = 1.72 × 10−8 Ω𝑚

20 − 𝑔𝑎𝑢𝑔𝑒:𝑅

𝐿=1.72 × 10−8 Ω𝑚

5.2 × 10−7 𝑚2= 0.033 Ω/𝑚

16 − 𝑔𝑎𝑢𝑔𝑒:𝑅

𝐿=1.72 × 10−8 Ω𝑚

13 × 10−7 𝑚2= 0.013 Ω/𝑚

20 − 𝑔𝑎𝑢𝑔𝑒 ℎ𝑎𝑠 𝑎𝑏𝑜𝑢𝑡 3 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

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Find new resistivity

𝜌 = 4 × 10−5 Ω𝑚 1 + 3 × 10−21

°𝐶350 °𝐶 − 200 °𝐶 = 2.2 × 10−4 Ω𝑚

Find resistance

𝑅 =𝜌𝐿

𝐴=

2.2 × 10−4 Ω𝑚 1.3𝑚

2 × 10−7 𝑚2= 1430 Ω

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Current in some superconductors have be constant for many years

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V=IR I = V/R

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Power

𝑃 =𝑉2

𝑅=

120 𝑉 2

1430 Ω= 10.1 𝑊

Energy use

𝑃 =𝑊

𝑡𝑊 = 𝑃𝑡 = 10.1 𝑊 86400 𝑠 = 872640 𝐽

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𝐸 = 0.0101 𝑘𝑊 720 ℎ = 7.272 𝑘𝑊ℎ𝐶𝑜𝑠𝑡 = 7.272 𝑘𝑊ℎ $0.15 = $1.09

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I0 and V0 stand for the maximum value

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𝑉𝑟𝑚𝑠 =156 𝑉

2= 110 𝑉

𝐼𝑟𝑚𝑠: 𝑃 = 𝐼𝑉 → 60𝑊 = 𝐼 110 𝑉 → 𝐼𝑟𝑚𝑠 = 0.55 𝐴

𝑃 =𝑉2

𝑅→ 60𝑊 =

110 𝑉 2

𝑅→ 𝑅 =

110 𝑉 2

60𝑊→ 202 Ω

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Wire resistance varies directly with L and inversely with A

If you use an extension cord, use one with thick wires and short length to reduce resistanceRemember small gauge means big wires

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Thin wires have higher R than thick wires

Heat can’t escape from coiled wires and they melt

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Rs is the equivalent resistance in Series

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Circuit board and multimeter to measure

5.17 k + 10.09 k = 15.26 k

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𝑅𝑆 = 3 8 Ω + 12 Ω = 36 Ω𝑉 = 𝐼𝑅 → 120 𝑉 = 𝐼 36 Ω 𝐼 = 3.33 𝐴𝑉 = 𝐼𝑅 𝑉 = 3.33 𝐴 8 Ω = 26.7 𝑉

𝑉 = 3.33 𝐴 12 Ω = 40 𝑉𝑃 = 𝐼2𝑅 𝑃 = 3.33 𝐴 2 8 Ω = 88.9 𝑊 𝑃 = 3.33 𝐴 2 12 Ω = 133.3 𝑊

𝑃𝑡𝑜𝑡𝑎𝑙 = 3 88.9 𝑊 + 133.3 𝑊 = 400𝑊 𝑃 = 𝐼2𝑅 = 3.33 𝐴 2 36 Ω = 400𝑊

Notice you can total the power and the voltage

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1

𝑅𝑃=

1

𝑅1+

1

𝑅2

1/𝑅𝑃 =1

1004 Ω+

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101 Ω= 0.000996/Ω + 0.00990/Ω = 0.010897/Ω

𝑅𝑃 =1

0.010897/Ω= 91.8 Ω

𝑉 = 𝐼𝑅 → 3 𝑉 = 𝐼 91.8 Ω 𝐼 = 0.0327 𝐴 = 32.7 𝑚𝐴

𝑉 = 𝐼𝑅 3 𝑉 = 𝐼 1004 Ω 𝐼 = 0.0030 𝐴𝑉 = 𝐼𝑅 3 𝑉 = 𝐼 101 Ω 𝐼 = 0.0297 𝐴

𝐴𝑑𝑑 𝑡ℎ𝑒𝑚 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 0.0327 𝐴

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𝐶𝑜𝑚𝑏𝑖𝑛𝑒 𝑓𝑎𝑟 𝑙𝑒𝑓𝑡 𝑏𝑟𝑎𝑛𝑐ℎ 𝑠𝑒𝑟𝑖𝑒𝑠 → 10090 Ω + 5170 Ω = 15260 Ω

𝐶𝑜𝑚𝑏𝑖𝑛𝑒 𝑙𝑒𝑓𝑡 𝑡𝑤𝑜 𝑏𝑟𝑎𝑛𝑐ℎ𝑒𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 →1

𝑅=

1

15260 Ω+

1

100900 Ω→

1

𝑅

= 7.54 × 10−5Ω→ 𝑅 = 13255 Ω

𝑇ℎ𝑒 𝑟𝑒𝑠𝑡 𝑖𝑠 𝑠𝑒𝑟𝑖𝑒𝑠 → 13255 Ω + 1004 Ω + 101 Ω = 𝟏𝟒𝟑𝟔𝟎 𝛀𝑉 = 𝐼𝑅 → 3 𝑉 = 𝐼 14360 Ω → 𝐼 = 2.09 × 10−4 𝐴 = 209 𝑚𝐴

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Far left two branches (parallel): 1

𝑅=

1

1004 Ω+

1

100900 Ω→ 𝑅 = 994.1 Ω

Combine series: 𝑅 = 994.1 Ω + 5170 Ω = 6164.1 Ω

Combine parallel:1

𝑅=

1

6164.1 Ω+

1

10090 Ω→ 𝑅 = 3826.5 Ω

Combine series: 𝑅 = 3826.5 Ω + 101 Ω = 3927 Ω

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𝑉 = 𝐼𝑅 (𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑤/𝑜 𝑏𝑎𝑡𝑡𝑒𝑟𝑦)2 𝑉 = 𝐼(20 × 10 Ω) → 𝐼 = 0.01 𝐴

𝑉 = 𝐼𝑅 (𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒)𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑟𝑜𝑝 𝑎𝑐𝑟𝑜𝑠𝑠 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

3 𝑉 − 2 𝑉 = 1 𝑉1 𝑉 = 0.01 𝐴 𝑅 → 100 Ω = 𝑅

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Combine parallel circuits1

𝑅= 5

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15 Ω→ 𝑅 = 3 Ω

Combine with internal resistance𝑅 = 3.02 Ω

Find current draw𝑉 = 𝐼𝑅

1.5 𝑉 = 𝐼 3.02 Ω → 𝐼 = 0.497 𝐴Use the circuit w/o battery to find terminal voltage

𝑉 = 𝐼𝑅𝑉 = 0.496 𝐴 3 Ω = 1.49 𝑉

Think of resisters in series and parallel

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Loop Rule (starting top left going CCW)𝐼 10090 + 4.5 𝑉 + 𝐼 5170 + 𝐼 101 + 𝐼 1004 = 3 𝑉

16365 Ω 𝐼 + 4.5 𝑉 = 3 V16365 Ω 𝐼 = −1.5 𝑉

𝐼 = −9.17 × 10−5 𝐴 = 91.7 𝜇𝐴

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𝐿𝑒𝑓𝑡 𝐽𝑢𝑛𝑐𝑡𝑖𝑜𝑛: 𝐼3 = 𝐼1 + 𝐼2𝑇𝑜𝑝 𝐿𝑜𝑜𝑝 𝐶𝐶𝑊: 3 𝑉 = 𝐼2 1004 + 𝐼2 101 − 𝐼1 100900

𝐵𝑜𝑡𝑡𝑜𝑚 𝐿𝑜𝑜𝑝 𝐶𝑊: 4.5 𝑉 + 3 𝑉= 𝐼3 10090 + 𝐼3 5170 + 𝐼2 1004 + 𝐼2 101

𝑆𝑦𝑠𝑡𝑒𝑚𝐼1 + 𝐼2 − 𝐼3 = 0

−100900 Ω 𝐼1 + 1105 Ω 𝐼2 = 3 𝑉1105 Ω 𝐼2 + 15260 Ω 𝐼3 = 7.5 V

𝐼1 = −2.45 × 10−5 𝐴, 𝐼2 = 4.81 × 10−4 𝐴, 𝐼3 = 4.57 × 10−4 𝐴

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Made of magnets, wire coil, spring, pointer and calibrated scale.

Current flowing through the coil makes it magnetic, so it wants to move. The stronger the current the more the coil will rotate.

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Example of Shunt resistors•Want to measure 100 mA, but meter’s coil only reads 0.100 mA. •Have shunt resistor take 99.9 mA and the coil only gets .1 mA•To know how big to make the shunt resistors, the resistance of the coil needs to be known.

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Large resistor is added because if V is constant Big R means small I

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Current no longer flows because the parallel plates aren’t connected and it can’t accept anymore charge

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Draw the circuit with Battery, capacitor, and resistor

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𝑇𝑖𝑚𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡: 𝜏 = 𝑅𝐶 = 800000 0.000005 𝐹 = 4 𝑠𝑀𝑎𝑥 𝐶ℎ𝑎𝑟𝑔𝑒: 𝑄 = 𝐶𝑉 = 0.000005 𝐹 12 𝑉 = 0.000060 𝐶 = 60 𝜇𝐶

𝑀𝑎𝑥 𝐶𝑢𝑟𝑟𝑒𝑛𝑡: 𝐼 =𝑉

𝑅=

12 𝑉

800000= 0.000015 𝐴 = 15 𝜇𝐴

𝐶ℎ𝑎𝑟𝑔𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛: 𝑞(𝑡) = 60 1 − 𝑒−𝑡4 𝜇𝐶

𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛: 𝐼(𝑡) = 15𝑒−𝑡4 𝜇𝐴

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