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## Intro to Comp Apps & Concepts

by: Mrs. Gonzalo Renner

27

0

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# Intro to Comp Apps & Concepts IS 100

Marketplace > Mohawk Valley Community College > Science > IS 100 > Intro to Comp Apps Concepts
Mrs. Gonzalo Renner

GPA 3.83

Staff

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COURSE
PROF.
Staff
TYPE
Study Guide
PAGES
0
WORDS
KARMA
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## Popular in Science

This 0 page Study Guide was uploaded by Mrs. Gonzalo Renner on Sunday November 1, 2015. The Study Guide belongs to IS 100 at Mohawk Valley Community College taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/233236/is-100-mohawk-valley-community-college in Science at Mohawk Valley Community College.

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Date Created: 11/01/15
Circuits 1 Practice Work 1 A ashlight uses two 15 volt batteries in series with a single bulb If the current drawn by the bulb is 50 mA determine A The effective resistance of the bulb B The power dissipated by the bulb C The life of the batteries if they are rated at 1 amphour 2 I like toasted bagels It takes about 4 minutes to properly toast a bagel in my toaster oven The oven is rated at 1500 watts If NiMo charges me 11 cents per KWH determine A The energy used to toast the bagel in KWH B The yearly energy cost to toast one bagel per day C The current drawn by the oven 3 A certain chunk of material has a resistance of 100 Ohms Determine the new resistance if A The length is doubled B The area is doubled C The length and area are both tripled D The material is altered so that its resistivity is increased ten fold 4 An audio ampli er has an ef ciency of 50 The loudspeaker it drives has an ef ciency of 10 A What is the net ef ciency B For a loudspeaker output of 2 watts what power must be drawn by the ampli er 5 A 60 volt source feeds four series connected resistors R1 is 100 Ohms R3 is 500 Ohms R4 is 200 Ohms R4 also dissipates 5 watts Determine A The value of R2 B The voltage drop across each resistor C The total power dissipated in the circuit resistors Circuits 1 Practice Work Answers 1 A RVI 3 V50 mA 60 Ohms B PIV 50mA3V150mW C 1 Ah50 mA 20 hours 2 4 minutes 460 0667 hours A 0667 hours 1500 W 100 WH lKWH B 365 daysyear 1 KWHday 365 KWHyear At 11KWH cost 401 C IPV 1500 W120 V 125 amps 3 Remember Rp lengtharea A R is doubled to 200 Ohms B R is halved to 50 Ohms C No change still 100 Ohms D R is 1k Ohms 4 A mean n2 111675 1 05 ie 5 B In other words 2 watts is 5 of what Pin 2 W ln Pin 2 W 105 Pin40 W 5 Knowing R4 and PR5 you can nd I PIZR or IVPR IV5 W200 Ohms 150 mA You can now nd the drops on R1 R3 R4 VR1IR1 VR150 mA 100 Ohms VR15V VmIR3 Vm50 mA 500 Ohms Vm25V VR4IR4 VR450 mA 200 Ohms VR410V From KVL sum of rises must equal sum of drops so VR2E VR4 VR4 VR4 VR260V 7 5V 7 25V 7 10V VR220V You now know the drop across R2 and the current through it so R2 VRzI R2 20 V50 mA R2 400 Ohms The total power is found using the total voltage applied and the total current drawn PIV P50mA60V P3W

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