Can someone please show me how to go about doing the following problem?

Consider the group of resistors shown in Figure 21-37, where R1 = 4.0 and = 9.1 V.

Find the current in each resistor.

And determine the potential difference between points A and B.

Thank you!

Kirchoff rules:

1.  The sum of the currents going into a junction is zero

2.  The sum of potentials in any closed circuit is zero.

Now we have to see how we go about implementing them. For this we need to use conventions.

The first convention is: a current going into a junction is positive, while a current coming out is negative.

In the junction above, I1 is going in therefore in the junction equation it will have a positive sign, while I2 and I3 will have negative signs, since they are coming out. So according to the first rule:

This is simply the rule “what comes in must come out”.

The second convention is that if the direction loop exits the positive end of the battery, the battery potential in the equation is positive. The sign potential on a resistor is determined by the direction of the current through the resistor with respect to the direction of the loop. If the current goes with the loop, the resistor potential IR is negative, if it goes against the loop it the potential sign in the equation is positive.

Let’s apply these rules to a very simple circuit: one battery and one resistor.

The first step is to choose (arbitrarily) the direction of the loop:

Now, the second step is to mark a negative sign near the entry point of the loop into the resistor, and a positive sign next to the exit point:


Now we choose the direction of the current. This choice is also arbitrary. For the sake of demonstration the principle, let’s choose the “wrong” direction:

Now we have to apply the conventions to the second Kirchoff rule.

The loop exits the positive end of the battery, so V has a positive sign in the equation. The current I enters the resistor against the loop, so the potential on the resistor IR is also positive (that is why it is useful to mark +/- next to the resistor – It reminds us what sign the potential should have).

So the sum of all the potentials along the loop is zero so we get:

Rearranging we get for the current:

This is Ohm’s law, only with a negative sign. No need to panic. A negative current just means that the direction we chose for this particular current should be reversed.

So this is how to solve circuits with Kirchoff rules. Let’s go now to the assignment.

The first step is to choose the direction of the loops. We can choose any closed circuit for a loop.

Here I chose the obvious loops. However, we could have chosen the outer circuit:

And we would get the same results.


The next step is to mark the entry and exit points of the resistors:

Note that we marked R3 twice, since we have two loops going through it. Each loop direction determines the potential sign when we write the loop equation.

The next step is to mark al the currents:

Note that I chose a direction which at the end will come out reversed (I3). This is useful as a checkpoint. When we calculate the currents we expect this current o come out negative. If it doesn’t – we probably made a mistake somewhere. However, be careful with this trick. Use it only when you are absolutely certain in the final direction of the current. It is usually safe to assume that the largest power source in the circuit pushes the current from its positive terminal.

Now we are ready to write the equations.

The junction equation is (all the currents are going into the junction):

The left loop:

Note that the current through R4 is the same as the current that goes out of V1 (what goes out of a battery must return to it), therefore the current through R4 is I1.

For the right loop:

Therefore we have three equations with three unknowns (the currents) that we need to solve for.

Or, rearranging yet again:

Plugging in the numbers:

From the first equation:

Substituting this in the other two equations:

Multiplying the top equation by 6.7 and the bottom equation by 13.8 we get:

Note the negative sign, it means that we chose a current direction which is opposing the real current’s direction.

.

Plugging it into the bottom equation we obtain:

And from the junction equation:

So to summarize:

So the final circuit looks like:

Note that point B is at the same potential as point C since they are connected by a conductor. Therefore the potential difference between point A and point B is the same as the potential difference between point A and point C. But the potential difference between point A and point C is the potential difference across R3, and according to Ohms law it is:

The potential across R2 is

And together:

Which is a nice confirmation that we probably got the right result.