# Homework Help: Basic circuit current calculation

1. Nov 8, 2012

### letsfailsafe

1. The problem statement, all variables and given/known data
Required data provided on the image

2. Relevant equations
Not required

3. The attempt at a solution
I can easly calculate the voltage but no idea how to calculate current...

Here is the image:

Many thanks

2. Nov 8, 2012

### CWatters

I'm not familiar with the use of that symbol or notation.

In the first problem you say "DEFG = 2V". The circuit suggests the voltage across D to G is 8V. So you must mean each one is 2V.

But then in problem 2 you say "CD = 8V". The circuit appeard to suggest the voltage across C to D is 8V. So you must mean they sum to 8V.

3. Nov 8, 2012

### CWatters

Are they light bulbs?

4. Nov 8, 2012

### letsfailsafe

Sorry for my notations...

DEFG| 2V means D, E, F, G, all have 2V

of course CD| 8V means that C and D both have 8V

Thought I should save some time and space by putting them all together...

5. Nov 8, 2012

### CWatters

In the second problem (eg No. 6) if B = 8V how can C = D = 8V ?

Otherwise apply KVL around the loop to work out the voltage across EF etc

Last edited: Nov 8, 2012
6. Nov 8, 2012

### CWatters

In the first problem you can't work out the current for each of the three branches without knowing more about the parts. Is there some text on the previous page?

7. Nov 8, 2012

### letsfailsafe

The question:
"In each circuit the lamps are identical. Calculate the current through, and voltage across, each lamp."

For first problem (question 5).
A: 18V, 6A
B,C: 4V 2A
D,E,F,G: 2V 1A

For second problem (question 6).
A: 12V 6A
B: 8V 4A
C,D: 4V 2A
E,F,G,H,I,J: 4V 2A

I've got most of the voltages right but I have no idea how they got those currents...

Last edited: Nov 8, 2012
8. Nov 9, 2012

### Minki

It looks like a ratio problem to me.

Question 5.

Those answers can't be right. From Kirchhoff's law of currents the sum of the currents in those three branches must be 7A.

By inspection the current ratios are:

A: 4
BC: 2
DEFG: 1

Sum of ratios = 7, 7A/7 = 1, so 1 is now the multiplier.

A: 4 x 1 = 4A
BC: 2 x 1 = 2A
DEFG: 1 x 1 = 1A

-----------------------------------------------------
Question 6 first network.

The current through the BCD set is 6A with current ratios:

B: 2
CD: 1

Sum of ratios = 3, 6A/3 = 2, so 2 is now the multiplier.

B: 2 x 2 = 4A
CD: 1 x 2 = 2A

----------------------------------------------------
Question 6 second network.

The current through the EFGHIJ set is 6A with current ratios:

EF: 1
GH: 1
IJ: 1

Sum of ratios = 3, 6A/3 = 2, so 2 is now the multiplier.

EF: 1 x 2 = 2A
GH: 1 x 2 = 2A
IJ: 1 x 2 = 2A

A is obviously 6A.

Hope this helps.

Last edited: Nov 9, 2012