Abbsdad

past paper 5.pdf

Hi,

Could someone help me with question 1b & 1c in the attached question paper.

Many thanks in advance.

ssylee

Which part of the question are you unclear about?

Abbsdad

Hi ssylee,

I am not sure on how to break the circuit into sections to carry out the superposition calculations.
Previous circuits have all been calculated using Kirchoff's Laws and also have had no more than three resistors.

Many thanks for any help or advice.

ssylee

For 1b, first short out the 10V source first, then draw your loops and find I due to the 15V source; next, leave the 10V source in the diagram, but this time short out the 15V source, then draw your loops and find I due to the 10 V source. Your final result should be I = I due to 15V source + I due to 10 V source.

For 1c, draw 3 loops of current, I1, I2, and I3, in cw or ccw direction (they should be in the same direction to avoid confusion). Use Kirchoff's Laws to lay out an equation for each loop. Then solve the 3 by 3 linear equation for the loop currents. For I, it should be the total amount of current flowing through that particular part, which is I1-I2 or I2-I1 depending on your directions. From that point on, you should be able to get I.

ljcox

I solved it using McMillman's Theorem.

It and Thevenin's Theorem are very useful.

__________________
Len

Abbsdad

Folks,

Many thanks.

Abbsdad

Folks,

Still having problems with this.
Just can't work out which way to arrange the loops.
Now starting to pull my hair out.

Thanks in advance.

ljcox

The Superposion solution is attached.

For 1 (c), you simply choose loops that are the most convenient.

I would use A, B, 0 for the I1 loop and C, B, 0 for the I2 loop.

The third loop A, C, 0 is for I3.

But forget about the I3 loop when solving for I1 & I2 as it is irrelevant.

Then you calculate I3 from the third loop,

Attached Images

Superposition.png (75.4 KB, 39 views)

__________________
Len

Abbsdad

Hi ljcox,

Many thanks for your reply.

Is the total current in the circuit 49.13 Amps.

Heres hoping.

ljcox

I afraid that's wrong. All you have to do to check your maths is to work backwards and check if Kirchoff's laws are satisfied, ie. use the result to check if it is correct.

40.13 A through a 2 Ohm resistor means there is 49.13 * 2 = 98.26 Volt across it. This can't be correct since the voltage supplies are only 10 and 15 Volt.

__________________
Len

Abbsdad

Hi again ljcox,

I have tried again and come up with an answer of 7.38 Amps.

Fingers crossed.

ljcox

7.38 * 2 = 14.76 wich means a current of about 30 mA through the 8 Ohm and a negative current through the n4 Ohm thus Kirchoff's current law is NOT satisfied.

Try again

__________________
Len