Calculating Current

[Dorian Gray]

Greetings Mathematicians,

I was wondering if someone could please help me calculate the current through the 55 ohm capacitor. I know that the entire current is 0.28 A. Could anyone please help me setup the calculation? Thanks for any and all help.
In the picture, The first resistor is 12 ohm, and the resistors in parallel are 55 and 70 ohm. A 12 V battery is used.
V=IR is a useful formula (volts=current * resistance)

 

[Deleted member 4993]

Greetings Mathematicians,

I was wondering if someone could please help me calculate the current through the 55 ohm capacitor. I know that the entire current is 0.28 A. Could anyone please help me setup the calculation? Thanks for any and all help.
In the picture, The first resistor is 12 ohm, and the resistors in parallel are 55 and 70 ohm. A 12 V battery is used.
View attachment 2598 V=IR is a useful formula (volts=current * resistance)

That cannot be correct!

Anyway, there are various ways to do this problem. You can calculate the Voltage between the left-hand-junction and B by calculating the voltage drop through the resistor at A.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[Dorian Gray]

That cannot be correct!

Anyway, there are various ways to do this problem. You can calculate the Voltage between the left-hand-junction and B by calculating the voltage drop through the resistor at A.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

Greetings Mathematicians,
I was wondering if someone could please help me with this problem

A 12-V battery is connected to terminals A and B in the figure. Take R=70omega

Part A. Find the current in the 12-omega resistor

MY WORK: V=IR
12= I(42.8)
0.28 Amps = I ******THIS PART IS CORRECT.**** IT IS COMPUTER GRADED

Part B. Find the current in the 55-omega resistor
*This is where I am stuck. I tried the following:
V=IR
12=I(55omega)
.22 Amps = I

THIS IS WRONG. My homework is computer graded, and I am only allowed 3 chances before I lose credit.

I do not know how to find the current in the 55omega branch. If someone can help me with that, then I will be able to answer part C


Part C. Find the current in the 70omega resistor
****If I know how to do part B then I will be able to solve for C as well

PartD. I don't know how to do this yet, but I am more concerned about parts b and c at the moment

 

[DrPhil]

Greetings Mathematicians,
I was wondering if someone could please help me with this problem

A 12-V battery is connected to terminals A and B in the figure. Take R=70omega

Part A. Find the current in the 12-omega resistor

MY WORK: V=IR
12= I(42.8) ?? Where did 42.8 come from?
0.28 Amps = I ******THIS PART IS CORRECT.**** IT IS COMPUTER GRADED

Part B. Find the current in the 55-omega resistor
*This is where I am stuck. I tried the following:
V=IR
12=I(55omega)
.22 Amps = I

THIS IS WRONG. My homework is computer graded, and I am only allowed 3 chances before I lose credit.

I do not know how to find the current in the 55omega branch. If someone can help me with that, then I will be able to answer part C


Part C. Find the current in the 70omega resistor
****If I know how to do part B then I will be able to solve for C as well

The only equation you need is V=IR, or R=V/I, or I=V/R

What you did wrong in the 2nd part was that the Voltage across the 50-ohm resistor is NOT 12V. You have to calculate the voltage drop across 12 ohms when current is 0.28A, then subtract that form 12V to find the voltage across the 55-ohm branch.

The same voltage is driving the current through the 70-ohm branch. As a check, the current through the parallel resistors had better add up the the total current of 0.28A !

 

[wjm11]

Greetings Mathematicians,
I was wondering if someone could please help me with this problem

A 12-V battery is connected to terminals A and B in the figure. Take R=70omega

Part A. Find the current in the 12-omega resistor

MY WORK: V=IR
12= I(42.8)
0.28 Amps = I ******THIS PART IS CORRECT.**** IT IS COMPUTER GRADED

Part B. Find the current in the 55-omega resistor
*This is where I am stuck. I tried the following:
V=IR
12=I(55omega)
.22 Amps = I

THIS IS WRONG. My homework is computer graded, and I am only allowed 3 chances before I lose credit.

I do not know how to find the current in the 55omega branch. If someone can help me with that, then I will be able to answer part C


Part C. Find the current in the 70omega resistor
****If I know how to do part B then I will be able to solve for C as well

PartD. View attachment 2599I don't know how to do this yet, but I am more concerned about parts b and c at the moment

42.8 ohms is the correct total resistance if R = 70 ohms. Therefore, your .28A current calculation is correct. Next you need to calculate voltage drops across the individual resistors. Since all of the current must pass across the 12 ohm resistor, the voltage drop across this element is v = IR = (.28A)(12 ohm) = 3.3645 V (approx.).

The remainder of the total 12V drop (between terminals A and B) had to occur across the parallel portion of the circuit. So the voltage drop across the parallel resistor section is 12 - 3.3645 = 8.8355 V. Use this voltage drop to calculate the current across the 55 ohm and 70 ohm resistors. The two currents must add up to the .28 A calculated previously. Does this make sense?

 

[Dorian Gray]

42.8 ohms is the correct total resistance if R = 70 ohms. Therefore, your .28A current calculation is correct. Next you need to calculate voltage drops across the individual resistors. Since all of the current must pass across the 12 ohm resistor, the voltage drop across this element is v = IR = (.28A)(12 ohm) = 3.3645 V (approx.).

The remainder of the total 12V drop (between terminals A and B) had to occur across the parallel portion of the circuit. So the voltage drop across the parallel resistor section is 12 - 3.3645 = 8.8355 V. Use this voltage drop to calculate the current across the 55 ohm and 70 ohm resistors. The two currents must add up to the .28 A calculated previously. Does this make sense?

Thank you very very very much! I did not know how to calculate the voltage drop. Let me work those things out and I'll post back with my work in a few minutes for you to check please! I think I understand what you typed.

 

[Dorian Gray]

The only equation you need is V=IR, or R=V/I, or I=V/R

What you did wrong in the 2nd part was that the Voltage across the 50-ohm resistor is NOT 12V. You have to calculate the voltage drop across 12 ohms when current is 0.28A, then subtract that form 12V to find the voltage across the 55-ohm branch.

The same voltage is driving the current through the 70-ohm branch. As a check, the current through the parallel resistors had better add up the the total current of 0.28A !

Thank you drphil! I did not see your post at first for some reason....

 

[Dorian Gray]

Hey all,

Here is my work AND I got parts B and C correct! Thank you!!!!

 

[Dorian Gray]

Would 12 omega increase and 55 omege and R omega decrease? I am not exactly sure how i am supposed to analyze that question

 

[wjm11]

Would 12 omega increase and 55 omege and R omega decrease? I am not exactly sure how i am supposed to analyze that question

Try a thought experiment: What would happen if R increased to infinity? How much current would flow through R? What would the total resistance of the circuit be? What would the total current be? Where would that current have to flow?

BTW, the "omega" symbol simply means "ohms", the unit of resistance. You should say "ohms" instead of omega.

 

[Dorian Gray]

Try a thought experiment: What would happen if R increased to infinity? How much current would flow through R? What would the total resistance of the circuit be? What would the total current be? Where would that current have to flow?

BTW, the "omega" symbol simply means "ohms", the unit of resistance. You should say "ohms" instead of omega.

If R increased to infinity, wouldn't the current decrease because of V=IR? R would get really big so I would have to decrease?

The total resistance would be 12 + 1/(1/55+1/infinity)


I imagine that 12 ohms would increase and 55 ohms and R would decrease??????

 

[Dorian Gray]

If R increased to infinity, wouldn't the current decrease because of V=IR? R would get really big so I would have to decrease?

The total resistance would be 12 + 1/(1/55+1/infinity)


I imagine that 12 ohms would increase and 55 ohms and R would decrease??????

 

[Dorian Gray]

SORRY for the extra posts! Computer issues at the moment

 

[wjm11]

If R increased to infinity, wouldn't the current decrease because of V=IR? R would get really big so I would have to decrease?

The total resistance would be 12 + 1/(1/55+1/infinity)


I imagine that 12 ohms would increase and 55 ohms and R would decrease??????

Yes! But run the actual numbers to be sure (and prove it to yourself): 12 + 1/(1/55+1/infinity) + 12 + 1/(1/55+0) = 12 + 55 = 67 ohms total resistance. By changing R to infinite resistance, we have stopped all current flow across that resistor. This is the same as removing the resistor from the circuit entirely, leaving just a series circuit with the 55 ohm and 12 ohm resistors. What is the new current?

 

[Dorian Gray]

Yes! But run the actual numbers to be sure (and prove it to yourself): 12 + 1/(1/55+1/infinity) + 12 + 1/(1/55+0) = 12 + 55 = 67 ohms total resistance. By changing R to infinite resistance, we have stopped all current flow across that resistor. This is the same as removing the resistor from the circuit entirely, leaving just a series circuit with the 55 ohm and 12 ohm resistors. What is the new current?

The computer said 12 and R decrease while 55 increased, so I was wrong..............

 

[DrPhil]

The computer said 12 and R decrease while 55 increased, so I was wrong..............

Please use phrases like "the current through the 12-ohm resistor" - just saying 12 decreases doesn't make sense, since "12" is a constant.

Always put units (V, A, ohm) on every number - MUCH easier to keep track if you do.

Since 55 ohms and R are in parallel, you can expect that if less current flows through R then more might flow through the 55 ohms. The total resistance seen by the Voltage source is
\(\displaystyle \displaystyle R_{total} = (12 \Omega) + \dfrac{1}{\dfrac{1}{55 \Omega} + \dfrac{1}{R}}
= (12 \Omega) + \dfrac{(55\Omega) × R}{(55\Omega) + R}\)

Since the total increases when R is large, the total current will decrease. Thus it is true that the current through the 12-ohm resistor is LESS, but if you will put in real numbers you should find that the current through the 55-ohm resistor IS larger when the parallel branch has less current.

 

[Dorian Gray]

Please use phrases like "the current through the 12-ohm resistor" - just saying 12 decreases doesn't make sense, since "12" is a constant.

Always put units (V, A, ohm) on every number - MUCH easier to keep track if you do.

Since 55 ohms and R are in parallel, you can expect that if less current flows through R then more might flow through the 55 ohms. The total resistance seen by the Voltage source is
\(\displaystyle \displaystyle R_{total} = (12 \Omega) + \dfrac{1}{\dfrac{1}{55 \Omega} + \dfrac{1}{R}}
= (12 \Omega) + \dfrac{(55\Omega) × R}{(55\Omega) + R}\)

Since the total increases when R is large, the total current will decrease. Thus it is true that the current through the 12-ohm resistor is LESS, but if you will put in real numbers you should find that the current through the 55-ohm resistor IS larger when the parallel branch has less current.

Thank you for the suggestions. They are much appreciated.


By the way, is this your academic profile by chance?
http://www-chne.unm.edu/faculty/heintz/heintz.html