How Do You Solve These Complex Circuit Problems?

[clairez93]

Homework Statement

See image attached. Please ignore all work you see around the problem. :]
I need help with 9, 10, and 11 in this picture.

Homework Equations

V = IR

The Attempt at a Solution

9. $$R_{eq} = R + (\frac{1}{2R} + \frac{1}{2R})^{-1} = 2R = 24$$
$$V' = (0.5)(\frac{1}{2R} + \frac{1}{2R})^{-1} = 6 V$$ (this is the voltage drop across the parallel set of resistors of 2R and 2R)
$$I_{2} = I_{3} = \frac{6}{2R} = 0.25$$ (this is the current across each of the parallel branches, and they are only equal because they have the same resistance and voltage]
$$I_{1} = I_{2} + I_{3} = .25 + .25 = .5$$
$$V_{i} = I_{1}R = (.5)(12) = 6$$
$$E = V' + V_{i} = 6+6 = 12$$

Correct answer is actually 24 V.


10. $$R_{eq} = (\frac{1}{250} + \frac{1}{300})^{-1} = 136.364$$
$$V = IR_{eq}$$
$$24 = I(136.364)$$
$$I = 0.176$$
$$V' = (0.176)(136.364) = 24$$ (this part is weird, is it because the whole circuit's in parallel anyway so there's really no voltage drop across anything?)
$$I = \frac{V}{R} = \frac{24}{300} = 0.08 A$$

Correct answer is actually 40 mA.

11. Taking both the loops clockwise:
$$0 = -10I_{1} -20I_{2} - 5I_{1} + 50$$
$$0 = 20I_{2} - 10I_{3} - 40$$

Stuck from here.


Any help or pointers would be appreciated.

 

[tiny-tim]

Hi clairez93!

9
$$I_{2} = I_{3} = \frac{6}{2R} = 0.25$$ (this is the current across each of the parallel branches, and they are only equal because they have the same resistance and voltage]
$$I_{1} = I_{2} + I_{3} = .25 + .25 = .5$$

Yes, I₂ = I₃ = I₁/2,

but where does 6/2R come from?

Use I = I₃ = I₁/2 and apply Kirchhoff's rules to the outer circuit.

 

[clairez93]

I used I = V/R, and the voltage across it I thought was 6, and then the resistance is 2R, so I = 6/2R. Is that not correct?

 

[tiny-tim]

… the voltage across it I thought was 6 …

why?

 

[clairez93]

$$V' = (0.5)(\frac{1}{2R} + \frac{1}{2R})^{-1} = 6 V$$ (this is the voltage drop across the parallel set of resistors of 2R and 2R)

From this.

 

[tiny-tim]

(this is the voltage drop across the parallel set of resistors of 2R and 2R)

From this.

ah! but 0.5 only the current through one of the 2Rs

 

[clairez93]

Aha! I see! so then the voltage drop across is really 12 V.
And therefore I2 = 12 / 2R = 0.5 = I3
And therefore I1 = I2 + I3 = 1
And therefore the voltage across R is IR = 1*12 = 12.
And therefore E = 24!
Thanks!

Now i am still having troubles with 10 & 11.

 

[tiny-tim]

Now i am still having troubles with 10 & 11.

10: I make the same as you … but much moire quickly, just by looking at the right-hand loop

11: use I₁ = I₂ + I₃

 

[clairez93]

Are my loop equations correct? I just want to make sure before I solve.

 

[clairez93]

Okay, so my loop equations are wrong, I believe because I am getting odd numbers.

 

[tiny-tim]

Okay, so my loop equations are wrong, I believe because I am getting odd numbers.

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