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Current problem

I am trying to find the current through each cell.

The main reason why i cant solve it is that im not sure how many different loops there are in the curcuit.

I will attach a pcture below
Reply 1
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Shaanv
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image.jpg
Original post by Shaanv
I am trying to find the current through each cell.

The main reason why i cant solve it is that im not sure how many different loops there are in the curcuit.

I will attach a pcture below

Try redrawing the circuit to look like something you may be more familiar with: i.e. the batteries placed to the left and right hand side of the diagram.

It should make the solution much clearer.

HINT: There are two loops.
Reply 3
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Shaanv
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For the loop from the 6v battery to the 2v battery, is the 2v battery just a 2v potenial drop as it is in the reverse direction to the 6v battery.

Thanks for ur help. Its much appreciated as always, enjoy the rep my good sir.
Original post by Shaanv
For the loop from the 6v battery to the 2v battery, is the 2v battery just a 2v potenial drop as it is in the reverse direction to the 6v battery.



Ahhhh. Not quite.

Using Kirchoff's rules we end up with three equations:

One current (KCL) equation and two voltage equations (KVL) which are then solved simultaneously.

Check out the link which guides you through the process step by step:

https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/branch-current-method/


Steps to follow for the “Branch Current” method of analysis:

(1) Choose a node and assume directions of currents.
(2) Write a KCL equation relating currents at the node.
(3) Label resistor voltage drop polarities based on assumed currents.
(4) Write KVL equations for each loop of the circuit, substituting the product IR for E in each resistor term of the equations.
(5) Solve for unknown branch currents (simultaneous equations).
(6) If any solution is negative, then the assumed direction of current for that solution is wrong!
(7) Solve for voltage drops across all resistors (E=IR).
(edited 7 years ago)
Reply 5
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Shaanv
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image.jpgI had a go at the problem.

I considered the current at the junction where the circuit branches to the 2v battery.

I then considered the two loops and i took the potential drops equal to the emfs of the batteries.

Then i rearranged equations and substituted the correct values in to work out the current passing the two cells as stated in the original question.

If u could take the time to have a look and let me know how i got on it would be much appreciated.
Reply 6
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Shaanv
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@uberteknik
Could u have a look at my calculations and see if they are along the right track plz.

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