Calculate the power dissipated in each of the three resistors shown below?
Volts is 9v, R1=100 ohms, R2=330ohms, R3=560 ohms, R3 and R2 are in parrarel while R1 is in series with both R2 and R3
Answers:
9 volts in circuit.
R1=100 ohms
R2=330 ohms
R3=560 ohms
you find what the total resistance of the aprallel resistors is first using the equation
total resistance in parallel=(R2*R3)/(R2+R3)
(you work out the values in the brackets first.)
you then get the answer of 207.6404494 we will call this value R9 for the next equation. (the value i have used only sperates it from other values. It is not a physics rule.)
you then use the equation for a resistor in series of total resistance=R1+R9
You then get the answer of 307.6404494
to work out what voltage si dissipated at these resistors you need to work out the share of volts each one as.
to do this, do 9 (volts) /total of series resistors to work out one share. then times the answer by the value of each resistor.
so R1 as a share of 2.93volts (to 2 decimal places.)
you need to now work out the share for the parallel resistors.
to do this first work out the total resistance of them*one share.
this then gives 6.074506939 volts
i thought i could do the second bit, but i have gone wrong somewhere. i have included my source for the first bit so you can have a read of it yourself if you wish, if it is school work and is not in for a while, i suggest you go onto bbc bitesize and click on sos teacher i think. you can then ask them the question and they will work out the answer with the woorking.
hope it works
sorry i cant be of any more help
560 in parallel with 330 = ~ 200 ohms
total resistance = ~ 300 ohms
current = 30 ma
power dissapated = IV = ~ 270 mw
power will be 90 mw across r1
110 mw across r2
60mw across r3
in real life it is perfectly acceptabel to use approximate values..most resistors are 5% tolerance and even good resistors have 1% or 2% tolerance
Essex Friendly has it right. I worked to a slightly greater tolerance and obtained the following:
R1 856mw
R2 112mw
R3 659mw
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Answers:
9 volts in circuit.
R1=100 ohms
R2=330 ohms
R3=560 ohms
you find what the total resistance of the aprallel resistors is first using the equation
total resistance in parallel=(R2*R3)/(R2+R3)
(you work out the values in the brackets first.)
you then get the answer of 207.6404494 we will call this value R9 for the next equation. (the value i have used only sperates it from other values. It is not a physics rule.)
you then use the equation for a resistor in series of total resistance=R1+R9
You then get the answer of 307.6404494
to work out what voltage si dissipated at these resistors you need to work out the share of volts each one as.
to do this, do 9 (volts) /total of series resistors to work out one share. then times the answer by the value of each resistor.
so R1 as a share of 2.93volts (to 2 decimal places.)
you need to now work out the share for the parallel resistors.
to do this first work out the total resistance of them*one share.
this then gives 6.074506939 volts
i thought i could do the second bit, but i have gone wrong somewhere. i have included my source for the first bit so you can have a read of it yourself if you wish, if it is school work and is not in for a while, i suggest you go onto bbc bitesize and click on sos teacher i think. you can then ask them the question and they will work out the answer with the woorking.
hope it works
sorry i cant be of any more help
560 in parallel with 330 = ~ 200 ohms
total resistance = ~ 300 ohms
current = 30 ma
power dissapated = IV = ~ 270 mw
power will be 90 mw across r1
110 mw across r2
60mw across r3
in real life it is perfectly acceptabel to use approximate values..most resistors are 5% tolerance and even good resistors have 1% or 2% tolerance
Essex Friendly has it right. I worked to a slightly greater tolerance and obtained the following:
R1 856mw
R2 112mw
R3 659mw
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