How Accurate is the Calculated Current I_b Using Nodal Analysis?

[Paymemoney]

Homework Statement

Determine I_b in the circuit below using nodal analysis.
http://img8.imageshack.us/img8/1339/circuit39.th.png

Homework Equations

Ohms Law

The Attempt at a Solution

Can someone tell me if i have done anything wrong in my first part of the question. Becuase i am pretty sure this is where i may have gone wrong.

Find the KCL for V_1

I_b = I_d + I_c

I_b = 12-V_1/250
I_c = V_1/50
I_d = V_1+60I_b/150

sub into KCL for V_1 to find the V_1 which i got as 1.28volts

then I_b as 42.88mA the answer is 39.67mA

P.S

 

[gneill]

Could you possibly supply a larger image? My microscope is in the shop.

 

[Paymemoney]

oops srry

heres a larger picture

http://img8.imageshack.us/i/circuit39.png/

 

[gneill]

I think that your expression for I_d should read:

I_d = (V_1 - 60I_b)/150

given the polarity shown for the controlled voltage source in the (thankfully) larger picture.

 

[Paymemoney]

ok i got two more question on nodal analysis

Qns 1:

Homework Statement

Calculate V_1 and V_2 in the circuit below using nodal analysis.

http://img30.imageshack.us/img30/4971/circuitqns.png

Homework Equations

Ohms Law

The Attempt at a Solution

Firstly i used KCL for supernode which gave me: i_1 + i_2 = i_3

i_0 = V_1/8
i_1 = V_2/4
i_2 = 3A

then i used KVL for the supernode which is the part i am unsure about. This is the equation i got from it but my final answer for V_2 was incorrect.

2 + 2 + V_2 - V_1 = 0
4 + V_2 - V_1 = 0
V_1 = 4 + V_2

Qns 2:

Homework Statement

Apply nodal analysis to find i_0 and the power dissipated in each resistor in the circuit below.

http://img30.imageshack.us/img30/4971/circuitqns.png

Homework Equations

Ohms Law

The Attempt at a Solution

Firstly i used KCL for supernode which gave me i_0 + i_1 + i_4 = -2

i_0 = V_1/6
i_1 = V_2/5
i_2 = 4A
i_3 = 2A
i_4 = V_2-V_3/3

then i used KVL for supernode loop1 : 10 + V_2 - V_1 = 0

then i used KVL for supernode loop2 : V_x + (V_2 - V_3) - 10 =0 where V_x = V_1 - V_3

so i get V_1 - V_3 + V_2 - V_3 - 10 = 0

V_1 -2V_3 + V_2 = 10

Can someone tell me if my main equation are correct becuase i am getting the incorrect answers.

P.S

 

[gneill]

Voltage changes across resistors are caused by currents through them. Voltage changes across fixed supplies are, well, fixed. Your equations need to reflect this. So, for example, in the first problem when writing the KVL for the branch between the nodes designated V2 and V1, you should have V1 = V2 + 2V - i1*2Ω. The 2V represents the fixed 2V supply.

 

[Paymemoney]

isn't i1*2 ohms positive why would i be negative? Because if you start from the V_1 point and go around clockwise you get 2i_1 + 2 + V_2 = V_1

 

[Paymemoney]

for question 2 would it be i_3 - i_4 * 3ohms = 10

how would you represent the voltage across i3? Because i just made it the current 2.

 

[gneill]

isn't i1*2 ohms positive why would i be negative? Because if you start from the V_1 point and go around clockwise you get 2i_1 + 2 + V_2 = V_1

i1 has its direction fixed in the diagram. There's no going clockwise if the current is given as counterclockwise. Or, if you're going to change it, you have to be sure to change it for every equation you write (especially current summations).

 

[Paymemoney]

for question 2 would it be i_3 - i_4 * 3ohms = 10

how would you represent the voltage across i3? Because i just made it the current 2.

anyone answer this question for me?

 

[gneill]

I'm afraid I don't understand the question. The current through the 3 Ohm resistor would appear to be i3 + i2. The diagram doesn't show an i4; Is it supposed to be the current through the 3 Ohm resistor?

Are you asking how you would indicate the voltage across the 2A current generator that is producing i3? Just put a directed arrow across it and label it with the voltage. If you're asking how to calculate it, then it's the sum of the 10V supply and the voltage across the 3 ohm resistor.

 

[Paymemoney]

yes that was what i meant. So i made the equation like the following but it doesn't give me correct answer? 10 + ((V_2 - V_3) / 3) = 0

 

[gneill]

yes that was what i meant. So i made the equation like the following but it doesn't give me correct answer? 10 + ((V_2 - V_3) / 3) = 0

So you're saying that you have found that V2 = V3? That shouldn't be.

What values did you find for i0 and i1?

 

[Paymemoney]

ok my first equation:

21V1 - 10V3 = -110 is derived from i0+i1+i4 = -2 the supernode

the other equation i used is node V3 i4 = -7 which comes out as V2 - V3 = -21

now i solved the simultaneous equation and i got:

1) 21V2 - 10V3 = -110
2) V2 - V3 = -21 *10

3) 10V2 - V3 = -210

1) - 3)
V2 = (-320/11)

therefore
10 + V2 = V1
V1 = -19
and i0 = V1/6 = -3.18A

 

[gneill]

I'm sorry Paymemoney, but I can't follow where your equations are coming from. I don't see a 21*V1 or a 10*V3 arising out of the given components. Here's how I went about it.

I began with noting that i4 is the sum of the 2A and 4A supplies (they join at the V3 node).
Then I noted that, for the supernode, (i3 + i2) - i1 - i0 - i3 = 0. Since i3 = 2A and i2 = 4A, that leaves

i1 = 4A - i0 ...(1)

Then,

V2 = i1*5Ω ...(2)
V1 = i0*6Ω ...(3)
V1 = V2 + 10V ...(4)

Since we're going for i0, I took the expressions above and evolved:

V1 = V2 + 10V
i0*6Ω = i1*5Ω + 10V ...substituting (2) and (3)
i0*6Ω = (4A - i0)*5Ω + 10V ...substituting (1)

Solving for i0 yields

i0 = 30V/11Ω = 2.727 A

Then, from (1),

i1 = 4A - i0 = 1.273 A

 

[Paymemoney]

the answer says that i0 = 29.45A

 

[gneill]

the answer says that i0 = 29.45A

I hate to argue with answer keys, but in this case I must.

All the currents shown are far less than 29 amps, so the only source for such a large current is the 10V supply. It could produce, at the very most, 10V/6Ω = 1.66A through the 6Ω resistor.

The result I gave for i0, 2.727A is far more likely! (it also has the benefit of being correct!)