Sarahisme Posted May 4, 2005 Share Posted May 4, 2005 hey could i please some help with the following questions? thanks Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 for part 1) is the voltage across each resistor just 9V?? Link to comment Share on other sites More sharing options...
Spyman Posted May 4, 2005 Share Posted May 4, 2005 for part 1) is the voltage across each resistor just 9V??The voltage across the circuit is 9 V. Voltage across each resistor is depending of the value of resistanse and the flow of amperes through each resistor. So voltage will differ between the resistors and all will be below 9V. Formulas: Ohm's law U(Volt)=R(Resistans)*I(Ampere) Resistanse in serie coupling Rtot = R1 + R2 + R3 ... and so on Resistanse in parallell Rtot = 1 / ( 1/R1 + 1/R2 + 1/R3 ... and so on ) Kirchoff's laws Sum of Ampere to and from any point is zero & Sum of Voltage through the circuit is zero Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 ok, how long will you be around for, i will try the questions now (using your above advice) and then if possible i could give some answers and you tell me if good or not? lol ?? Link to comment Share on other sites More sharing options...
klanger Posted May 4, 2005 Share Posted May 4, 2005 Spyman So the resistors in series are 112r + 110r = 222r of resistance? Those in parralel are 450r + 40r / 2 = 245r (or there abouts, I dont have a scientific calculator to get it spot on) and then? 222r + 245r + 10r (resistance on meter) = 477r ??? I think I went horribly wrong Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 ok i get total resistance of the system to be 222 ohms and then from that, the total current to be 9/222 = 4.05x10^-2 then from this work out that the voltage drop across the 112 ohm resistor is 4.54 V and the current through it is 4.05x10^-2 Amps and then the voltage through the 450 ohm resistor is 4.46 V and that the current through it is 9.91x10^-3 Amps and that the voltage through the 110 ohm resistor is 4.46V and the current through it is 4.05x10^-2 Amps and that the voltage through the 40ohm resistor is 4.46V and the current through it is Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 actually changed my mind, for part 1) i get 112ohm: current through: 4.01x10^-2 Amps voltage across: 4.49 Volts 450ohm: current through: 1.00x10^-2 Amps voltage across: 4.51 Volts 110ohm: current through: 3.01x10^-2 Amps voltage across: 3.31 Volts 40ohm: current through: 3.01x10^-2 Amps voltage across: 1.20 Volts hows that?!?! Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 anyone got any ideas for question 2 ???? ??? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 is the answer to 2) by nothing? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 actually i got b, what bout question 3??? Link to comment Share on other sites More sharing options...
klanger Posted May 4, 2005 Share Posted May 4, 2005 what is question 3? I only see two questions on your origional post Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 oops my bad, hang on.... Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 what is question 3? I only see two questions on your origional post 3. When measuring voltages, the resistance of the multimeter is 10 MΩ on all ranges. Would you expect this resistance to significantly affect the measurement of voltages in the above circuit? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 what i said was..... Since the resistance of the multimeter is so large, it would result in there being no voltage for the rest of the circuit beyond the multimeter. Link to comment Share on other sites More sharing options...
klanger Posted May 4, 2005 Share Posted May 4, 2005 I believe that is right yes, but are you sure the resistance on the multimeter is 10MΩ ? That seems extremely high to me, there is usually some resistance on the multimeter but not normally so much that it renders the meter useless. 10Ω would be more acceptable and though yeah it would effect the reading some, that fact that all resistors have verying tolerances would lose this resistance in the meter itself. What do you think? Link to comment Share on other sites More sharing options...
klanger Posted May 4, 2005 Share Posted May 4, 2005 on conventional resisters (four and five bands) the last band usually with a slighter larger gap between it and the preceding colour band, is the tolerance band. These are Brown = 1% Red = 2% Gold = 5% Silver = 10% This allows each resister to fluctuate by that %, so a 100Ω resister is Brown Black Black......Brown = 99Ω to 101Ω Brown Black Black......Red....= 98Ω to 102Ω Brown Black Black......Gold...= 95Ω to 105Ω Brown Black Black......Silver..= 90Ω to 110Ω Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 hmm ok i see what your saying, well i suppose since its not a "real physical" problem, i.e. its a theoretical circuit so yeah ? Link to comment Share on other sites More sharing options...
Spyman Posted May 4, 2005 Share Posted May 4, 2005 Hold on, I will edit my post whilst doing the calculations... Question 1: ========= 1) You must calculate the whole resistance of the circuit. Serie = 110 + 40 = 150 Ohm Parallell = 1 / (1/450 + 1/150) = 113 Ohm Serie = 112 + 113 = 225 Ohm 2) The flow of current from battery (I1) I1 = 9 / 225 = 40 mA 3) The voltage over 112 Ohm resistor (All flow goes through here) U(112) = 0.04 * 112 = 4.48 V 4) The voltage over 450 Ohm resistor (Sum in circuit is zero) U(450) = 9 - 4.48 = 4.52 V 5) The flow of current through 450 Ohm resistor (I2) I2 = 4.52 / 450 = 10 mA 6) The last flow of current (Sum of in connection point is zero) (I3) I3 = 40 - 10 = 30 mA 7) The voltage over 110 Ohm resistor U(110) = 0.03 * 110 = 3.3 V 8) The voltage over 40 Ohm resistor U(40) = 0.03 * 40 = 1.2 V Question 2: ========= When measuring the current with a multimeter You need to break up the circuit and insert the multimeter in serie with the component You are to measure the flow through. (Hence the low resistance inside the multimeter when measuring current.) Thus the new resistance of that part of the circuit will be: Serie = 110 + 40 +10 = 160 Ohm Then go through the parts in question 1 again down to point 6 (I3) and You will have a new value for I3. Question 3: ========= When measuring voltages, the resistance of the multimeter is very high, (10 MΩ seems good), because then You insert the multimeter as parallell. If measuring the voltage of a battery with low resistance You would drain it and if measuring a higher voltage the flow of current could be so high that the multimeter would melt. "Would you expect this resistance to significantly affect the measurement of voltages in the above circuit ?" On which resistor should a parallell resistance of 10 000 000 Ohm have the most effect and how much ? New resistance = 1 / (1/450 + 1/10000000) = 449.98 Ohm You can continue to calculate the voltage but I doubt i would change much. Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 Hold on' date=' I will edit my post whilst doing the calculations... Question 1: ========= 1) You must calculate the whole resistance of the circuit. Serie = 110 + 40 = 150 Ohm Parallell = 1 / (1/450 + 1/150) = 113 Ohm Serie = 112 + 113 = 225 Ohm 2) The flow of current from battery (I1) I1 = 9 / 225 = 40 mA 3) The voltage over 112 Ohm resistor (All flow goes through here) U(112) = 0.04 * 112 = 4.48 V 4) The voltage over 450 Ohm resistor (Sum in circuit is zero) U(450) = 9 - 4.48 = 4.52 V 5) The flow of current through 450 Ohm resistor (I2) I2 = 4.52 / 450 = 10 mA 6) The last flow of current (Sum of in connection point is zero) (I3) I3 = 40 - 10 = 30 mA 7) The voltage over 110 Ohm resistor U(110) = 0.03 * 110 = 3.3 V 8) The voltage over 40 Ohm resistor U(40) = 0.03 * 40 = 1.2 V Question 2: =========[/quote'] thanks Spyman, so i think i got questions 1 and 2 right. but could you please have a look at my answer to quesiton 3 (question 3 is posted a few posts down because i forgot to post it in the orginal thread) Thanks a million Spyman Sarah Link to comment Share on other sites More sharing options...
Spyman Posted May 4, 2005 Share Posted May 4, 2005 Sarah, You are welcome, The answer to Q3 in my above post #18. klanger, be very careful when using a multimeter without knowing the difference of measuring voltage and current. The multimeter can explode and cause serious injury ! If inserted wrongly in for instance the power jack in Your house, say 110 V AC and with only resistance of 10 Ohm, (switched for current), then the current through the meter would be I=U/R=110/10 = 11 Ampere and the power released as heat would be P=U*I=110*11=1210 Watt. Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 oh ok, thanks Spyman, sorry i didnt see it there before *embarrassed*, thanks again Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 lol, not to seem to thick, but i read through your answer to question 3, and i don't really get it completely..... damn my brain! "Would you expect this resistance to significantly affect the measurement of voltages in the above circuit?" is my answer to this quesiton right? its a few posts back , sorry for the hassling sarah Link to comment Share on other sites More sharing options...
Spyman Posted May 4, 2005 Share Posted May 4, 2005 Connect the multimeter as an parallell resistor to the resistor You want to measure the voltage over. 10 MOhm = 10 000 000 Ohm (M = Mega or Million) 10 MOhm parallell to 450 Ohm = 1/ (1/450 + 1/10000000) = 449.98 Ohm I2 = 10 mA gives without voltmeter = 0.01 * 450 = 4.5 V and with voltmeter = 0.01 * 449.98 = 4.4998 V To calculate the exact value You must do the whole calculation from the beginning again with the multimeter inserted, but as You can see in the approximated value the change is very small. If measuring over a resistor with smaller value the difference will be smaller, hence I picked the largest. Question: "Would you expect this resistance to significantly affect the measurement of voltages in the above circuit?" Your answer: "Since the resistance of the multimeter is so large, it would result in there being no voltage for the rest of the circuit beyond the multimeter." Your answer is not correct. Where do You connect the multimeter and how ? Maybe You are trying to insert it as when measuring currents ? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 4, 2005 Author Share Posted May 4, 2005 right ok, yeah thanks, that makes sense now Cheers Spyman Sarah Link to comment Share on other sites More sharing options...
Douglas Posted May 4, 2005 Share Posted May 4, 2005 hey could i please some help with the following questions? thanks total circuit resistance is......... 110 in series with 40 = 150 150 in parallel with 450 = 112.5 112.5 in series with 112 = 224.5 Ohms total current = 40 mils voltage drop across 112 = 4.5 V I'll do the rest if you want Link to comment Share on other sites More sharing options...
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