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Simple circuit diagram help

  • Engineering
  • Thread starter Asphyxiated
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  • #1
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Homework Statement



Calculate V1, i1, V2, and i2 in http://images3a.snapfish.com/232323232fp733;4>nu=52::>379>256>WSNRCG=3357955356347nu0mrj"

Homework Equations



Kirchhoff's Voltage Law (KVL) - The algebraic sum of voltages on a closed loop are equal to zero.
Kirchhoff's Current Law (KCL) - The algebraic sum of currents entering a node (or closed boundary) is zero.

Ohm's Law - V = IR, where V is voltage, I is current and R is resistance.

Various other things like how to add resistors in parallel and series to make a circuit simpler to understand. e.g. R1 and R2 in series are equal to

[tex] R_{eq}= R_{1}+R_{2} [/tex]

in parallel on the other hand

[tex] \frac {1}{R_{eq}}=\frac {1}{R_{1}}+\frac{1}{R_{2}} [/tex]

The Attempt at a Solution



So this is our first homework assignment and I am just looking for some guidance here. In the figure I linked to I labeled i0 so that I could easily reference it. This should be equal to 12/4 which is 3 amps correct? and

[tex] i_{0}=i_{1}+i_{2} [/tex]

yes? That would also make V1 3*4=12 yes?

Is it necessary to combine any of the resistors to finish the problem? Also because there is no element where i1 is, would this mean there is no current usage and i1 is zero?

Thanks for any help in advance
 
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Answers and Replies

  • #2
berkeman
Mentor
57,296
7,275

Homework Statement



Calculate V1, i1, V2, and i2 in http://images3a.snapfish.com/232323232fp733;4>nu=52::>379>256>WSNRCG=3357955356347nu0mrj"

Homework Equations



Kirchhoff's Voltage Law (KVL) - The algebraic sum of voltages on a closed loop are equal to zero.
Kirchhoff's Current Law (KCL) - The algebraic sum of currents entering a node (or closed boundary) is zero.

Ohm's Law - V = IR, where V is voltage, I is current and R is resistance.

Various other things like how to add resistors in parallel and series to make a circuit simpler to understand. e.g. R1 and R2 in series are equal to

[tex] R_{eq}= R_{1}+R_{2} [/tex]

in parallel on the other hand

[tex] \frac {1}{R_{eq}}=\frac {1}{R_{1}}+\frac{1}{R_{2}} [/tex]

The Attempt at a Solution



So this is our first homework assignment and I am just looking for some guidance here. In the figure I linked to I labeled i0 so that I could easily reference it. This should be equal to 12/4 which is 3 amps correct? and

[tex] i_{0}=i_{1}+i_{2} [/tex]

yes? That would also make V1 3*4=12 yes?

Is it necessary to combine any of the resistors to finish the problem? Also because there is no element where i1 is, would this mean there is no current usage and i1 is zero?

Thanks for any help in advance
There is a dead short across R2+R3. So what is the parallel combination of the dead short and R2+R3? So what does that leave you in the circuit?
 
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  • #3
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Well R2 and R3 are in series, i assume, no polarity sign was given for R2 so they should combine together to get 9 ohms and the definition of short circuit is a wire/connector with near perfect conductance or a resistivity near 0, right? So if the 9 ohm resistor are in parallel and you combine them 1/Req = 1/0+1/9.

I am not sure if the combined resistance does not exist or is it 9 ohms? (the reciprocal of 1/9, which would be the answer to 1/Req if you just leave out the undefined term/count as zero). Which will leave you with a 4 ohm resistor and 9 ohm resistor in series, am I going the right way?
 
  • #4
berkeman
Mentor
57,296
7,275
Well R2 and R3 are in series, i assume, no polarity sign was given for R2 so they should combine together to get 9 ohms and the definition of short circuit is a wire/connector with near perfect conductance or a resistivity near 0, right? So if the 9 ohm resistor are in parallel and you combine them 1/Req = 1/0+1/9.

I am not sure if the combined resistance does not exist or is it 9 ohms? (the reciprocal of 1/9, which would be the answer to 1/Req if you just leave out the undefined term/count as zero). Which will leave you with a 4 ohm resistor and 9 ohm resistor in series, am I going the right way?
It's better to avoid the 1/0 issue by re-writing the parallel combination equation:

[tex]R = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}[/tex]

And simplify the right hand side to get the fractions out of the denominator. What equation do you get for the equivalent parallel resistance then? And see how there are no 1/0 terms now when you substitute in your values....
 
  • #5
264
0
Well i simplified it to:

[tex] R_{eq}=\frac {R_{1}*R_{2}}{R_{2}+R_{1}} [/tex]

so plugging the numbers i get 0/9 so the combined resistivity is 0 and basically the only thing that matters is V1?

said another way, since there is a short circuit before R2 and R3 there will be no current flowing through them, right?
 
  • #6
berkeman
Mentor
57,296
7,275
Well i simplified it to:

[tex] R_{eq}=\frac {R_{1}*R_{2}}{R_{2}+R_{1}} [/tex]

so plugging the numbers i get 0/9 so the combined resistivity is 0 and basically the only thing that matters is V1?

said another way, since there is a short circuit before R2 and R3 there will be no current flowing through them, right?
Correct. All the current flows through the short circuit. Good!
 
  • #7
264
0
Ok so just to make sure then V1 is going to be 12v i1 is 3 amps and since there is no current through V2, it must be zero if i2 is 0, V=IR. Yeah?
 
  • #8
berkeman
Mentor
57,296
7,275
Ok so just to make sure then V1 is going to be 12v i1 is 3 amps and since there is no current through V2, it must be zero if i2 is 0, V=IR. Yeah?
Correct-a-mundo. :smile:
 
  • #9
264
0
thanks so much, i am flying through the rest, hope i don't jinx myself by saying this though!
 

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