Anyone who can help me on this dc circuits?

Question

the link for the images are as follows

http://allizione.multiply.com/photos/album/2/physics_help_me_dc_circuits

please please help me answer the question thanks! tell em also how to solve such. i just don't know if they are connected in series and parallel and why they are connected that way?

Answer 1

In Fig.1, this is neither a series nor a parallel circuit with the 5 ohm resistor in place.  The way to see what is going on with this kind of circuit is to pull out the 5 ohm resistor and see what is left.  What is left is the circuit in Fig.2.  Fig.2 is simply a series / parallel circuit.

First thing to do, with the 5 ohm resistor removed, look at the ratios between the 1 ohm and 3 ohm branch resistors (1:3) and the ratio of resistance in the other branch, whic is (2:4).  These ratios are not equal. This should tell you that there will be a voltage difference at the two points where the 5 ohm resistor is connected in the original circuit in Fig.1.  Ok, don't worry about that for the moment. The next thing you want to do is calculate the current flowing in each of the two resistive branches.  Fig.2 shows you that in one branch the series resistance is 1 ohm + 3 ohm = 4 ohms. Thus, the current flowing in that branch is: 10 volts / 4 ohms = 2500 mA, or 2.5 Amps.  OK, the voltage drop across the 1 ohm resistor is, 1 ohm x 2.5 Amps = 2.5 volts.  The voltage drop across the 3 ohms x (2.5 Amps) = 7.5 volts.

In the other branch, the current is, 10 volts / (2 ohms + 4 ohms) or, 10 volts / 6 ohms = 1.67 Amps.  The voltage drop across the 4 ohm resistor is: 4 ohms x 1.67 Amps = 6.68 volts.  The voltage drop across the 2 ohm resistor is, 2 ohms x 1.67 Amps = 3.34 volts.

Notice that the total voltage drops across each resistor in each of the two branches must add up to the supply voltage of 10 volts. If not, than there was an error in calculation.

Now, if you let the negative terminal of the battery be at circuit ground, the voltage drop across the 3 ohm resistor is 7.5 volts.  The voltage drop across the 4 ohm resistor is 6.68 volts.  The difference is 7.5v - 6.68v = 0.82 volts.  This is the voltage drop that would be seen across the 5 ohm resistor if it were in the circuit. So, placing the 5 ohm resistor back into the circuit, how much current would be flowing throug it?  Well, current is equal to E / R.   We know there is a voltage difference of 0.82 volts. Thus the current through the 5 ohm resistor is 0.82 v / 5 ohms = 164 mA or 0.164 Amps.

The resistive network is considered to be balanced. If you substract the voltage drops across the 1 ohm and 2 ohm resistors, they are the same as the substracted voltage drops across the 3 ohm and 4 ohm resistors.  This type of circuit is found in many balanced bridge devices such as RLC component measuring instruments, and in some forms of audio circuit oscillators.

The total resistance of the bridged resistor circuit is simply: 3 ohms + 1 ohm, in parallel with 4 ohms + 2 ohms. Total circuit resistance is: 4 ohms // 6 ohms =  2.4 ohms.

Check this: 10 volts / 2.4 ohms = 4.17 Amps.

That is approximately what you get when you add up the two branch currents, plus the small current flowing through the 5 ohm resistor.

Do you see that in a perfectly balanced resistive bridge network, it doesn't matter what resistor is used in place of the original 5 ohm resistor.  No current would flow in that resistor because the bridge is balanced and there would be zero voltage difference across that resistor. It is as if that resistance isn't there at all, but only when the network is perfectly balanced.

  

Answer 2

No 2

This is a series parallel combination.

1 and 3 in series make 4

2 and 4 in series make 6

4 and 6 are in parallel and equivalent is:

4*6/(4+6) = 2.4 ohms

Current in left leg is 10/4 = 2.5 amps

Current in right leg is 10/6 = 1.67 amps

Voltage across 1 ohm is 2.5*1 = 2.5v, which leaves 7.5v across 3 ohms

Voltage across 4 ohms is 1.67*4 = 6.67 v

voltage across 2 ohms is 1.67*2 = 3.33 v

No 1 is more complicated

and I'm out of time

.

Answer 3

Circuit 1:

Let's use some of the calculations made by Mr. billrussell42:

U3'=7.5V and U4'=6.667V (without R5=5 ohm).

We'll find I5 by using  the Thevenin's theorem:

1) The voltage difference between the points where 5 ohm resistor WILL be connected is E5=U3'-U4'=0.833V.

2) Then you must short-circuit the voltage source and find the equivalent resistance between the points where 5 ohm resistor will be connected, and it is:

Re=(1||3)+(2||4)=1*3/(1+3)+2*4/(2+4)=

3/4+4/3=25/12 ohm

3) Then the current through 5 ohm resistor is:

I5=E5/(Re+5)=(5/6)/(25/12+5)=

2/17~0.118 A,

and the voltage drop across it is

U5=5*I5=0.588 V

To find the currents and the voltage drops of the other resistors, one triangle  connection should be transformed to a star connection.

Let's transform the triangle, consisting of 1, 2 and 5 ohm resistors.

See the equivalent circuit:

http://i299.photobucket.com/albums/mm286...

R12=(1*2)/(1+2+5)=1/4=0.25 ohm

R15=(1*5)/(1+2+5)=5/8=0.625 ohm

R25=(2*5)/(1+2+5)=5/4=1.25 ohm

The equivalent resistance of the whole circuit is:

R=R12+(R15+3)||(R25+4)=

0.25+3.625||5.25=2.394 ohm

The current through the voltage source is:

I=U/R=10/2.394=4.176 A

Let R3=3 ohm and R4=4 ohm.

Then I3=I*(R4+R25)/(R3+R15+R4+R25)=

4.176*5.25/8.875=2.470 A

and I4=I-I3=4.176-2.470=1.706 A

The corresponding voltage drops are:

U3=R3*I3=7.41 V

U4=R4*I4=6.824 V

U5=U3-U4=0.586 V that confirms the previous calculations of U5.

Now back to the Circuit 1 (R5 is connected):

I1=I5+I3=2.588 A

I2=I4-I5=1.588 A

U1=1*I1=2.588 V

U2=2*I2=3.176 V

Check:

U=10V=U1+U3=9.998V

U=10V=U2+U4=10V

I=I1+I2=2.588+1.588=4.176A

Thevenin's theorem:

http://hyperphysics.phy-astr.gsu.edu/hba...

Triangle-star transformation:

http://en.wikipedia.org/wiki/Y-%CE%94_tr...