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[SOLVED] Circuits questions help(...supposedly they are easy.)
Here is a picture of the question with the diagrams
http://img153.imageshack.us/img153/9741/questioniy3.jpg
Resistors in series/parallel,kirchoff's laws.
For the part b) I just found the equivalent resistance and then used the fact that [itex]R_1=25R_2[/itex] and then found the total current using [itex]I=\frac{V}{R}[/itex]
and got:
[tex]I=(\frac{26}{26r+25R_2})*E[/tex]
so to find a desired value for I you would need to choose E,r and [itex]R_2[/itex] accordingly?
For part C) [This is where I think I did the question wrong]
p.d. through the voltmeter and [itex]60 \Omega[/itex] resistor =2V
So that the pd through the [itex]90 \Omega[/itex] resistor=6-2=4V
Current through the [itex]90 \Omega[/itex] resistor,I=[itex]\frac{4}{90}=.0444A[/itex]
Current through the [itex]60 \Omega[/itex] resistor,[itex]I_1[/itex]=[itex]\frac{2}{60}=0.033A[/itex]
By Kirchoff's first law:[itex]I=I_1 + I_2[/itex] ((I_2 is the current through the voltmeter))
so that [itex]I_2=I-I_1[/itex]
[tex]\frac{2}{R_v}=\frac{1}{90}[/tex]
and [itex]R_v=180 \Omega[/itex]
cii)
Total resistance=1038.46ohms
total current,I=6/1038.46=0.005777777~.0058A
pd through 900 ohm resistor=0.005777777*900=5.2V
pd through voltmeter and 600ohm resistor=6-5.2=0.8V
Not too sure what this says about voltmeters except that if they have resistance comparable to the other resistances they draw too much current and then some of the pd is lost in it.
Homework Statement
Here is a picture of the question with the diagrams
http://img153.imageshack.us/img153/9741/questioniy3.jpg
Homework Equations
Resistors in series/parallel,kirchoff's laws.
The Attempt at a Solution
For the part b) I just found the equivalent resistance and then used the fact that [itex]R_1=25R_2[/itex] and then found the total current using [itex]I=\frac{V}{R}[/itex]
and got:
[tex]I=(\frac{26}{26r+25R_2})*E[/tex]
so to find a desired value for I you would need to choose E,r and [itex]R_2[/itex] accordingly?
For part C) [This is where I think I did the question wrong]
p.d. through the voltmeter and [itex]60 \Omega[/itex] resistor =2V
So that the pd through the [itex]90 \Omega[/itex] resistor=6-2=4V
Current through the [itex]90 \Omega[/itex] resistor,I=[itex]\frac{4}{90}=.0444A[/itex]
Current through the [itex]60 \Omega[/itex] resistor,[itex]I_1[/itex]=[itex]\frac{2}{60}=0.033A[/itex]
By Kirchoff's first law:[itex]I=I_1 + I_2[/itex] ((I_2 is the current through the voltmeter))
so that [itex]I_2=I-I_1[/itex]
[tex]\frac{2}{R_v}=\frac{1}{90}[/tex]
and [itex]R_v=180 \Omega[/itex]
cii)
Total resistance=1038.46ohms
total current,I=6/1038.46=0.005777777~.0058A
pd through 900 ohm resistor=0.005777777*900=5.2V
pd through voltmeter and 600ohm resistor=6-5.2=0.8V
Not too sure what this says about voltmeters except that if they have resistance comparable to the other resistances they draw too much current and then some of the pd is lost in it.
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