# Find currents I1 and I2 in a 3-loop, 2-battery circuit

• prajakta
In summary, to find the currents I1 and I2 in the given circuit, we need to apply Kirchhoff's loop rule and use three loop equations. The currents through the 5 ohm and 8 ohm resistors cannot be assumed to be equal to I1 and I2 respectively. Negative current values are also acceptable, as they indicate that the actual current is flowing in the opposite direction of the assumed direction.
prajakta

## Homework Statement

Find currents I1 and I2 in the following circuit:

## Homework Equations

I know I have to use Kirchoff's loop rule but I am not sure of the current going through the 5 ohm and 8 ohm resistors, is it just I1 and I2 respectively? Also do I incorporate the large loop at the bottom?

## The Attempt at a Solution

-12 - (1)I1 - (5)I1 = 0
-9 - (1)I2 - (8)I2 = 0
I get negative answers with the above^

This is a practice midterm problem, so it's not my actual homework.[/B]

I haven't checked your arithmetic or if your stated equations are correct, but there is nothing wrong with getting negative values. Since these are linear equations, it only means that the direction of the current is actually going the other way as what was originally assumed.

prajakta said:
I know I have to use Kirchoff's loop rule but I am not sure of the current going through the 5 ohm and 8 ohm resistors, is it just I1 and I2 respectively? Also do I incorporate the large loop at the bottom?
As for the answer to that question, think of it this way. If the current going through the 5 ohm resistor were ##I_1##, the bottom loop, with the 10 ohm resistor, would not have any current going through it. Can you see why? Apply Kirchhoff's junction/node rule to junction on the left, and you'll see there have to be three currents... one going through the 10 ohm resistor, your ##I_1##, and whatever current is going through the 5 ohm resistor.

Hi prajakta,

Welcome to Physics Forums!

The currents I1 and I2 refer to just the currents in the components that they are next to. Other components will have different current values. Certainly the currents through the 5 and 8 Ohm resistors will not be equal to I1 and I2.

You can apply Kirchhoff's current law (KCL) throughout the circuit, defining "new" currents as required until every component is accounted for. Then you'll be in a position to use KVL to write loop equations and solve for all the currents.

Last edited:
scottdave
Sorry, my previous answer was considered too much help. Just know that you will need 3 loop equations, and all of the components need to be included at least once. Please note that I1 and I2 combine at the node, and the only think you know for sure is that all of this current will go to either the 8 ohm or the 5 ohm. Do not assume that the current through the 5 ohm is the same as i1.
As another poster stated, you can use I1 to represent the current through the 1ohm, but the current through the 5 ohm will be some combination of I1 and another current, from the 3rd loop. Similar with the 8ohm. I hope this helps. And yes, negative current values are OK. It just means that the actual current is flowing in the opposite direction of the assumed direction.

Last edited:

## 1. How do I determine the direction of the currents in the circuit?

The direction of the currents can be determined by applying Kirchhoff's current law, which states that the sum of currents entering a node must equal the sum of currents leaving the node. This means that the direction of the currents will depend on the orientation of the resistors and batteries in the circuit.

## 2. How do I calculate the total resistance in the circuit?

The total resistance in the circuit can be calculated by using Ohm's law, which states that resistance is equal to the voltage divided by the current. In this case, the voltage would be the sum of the two battery voltages, and the current would be the sum of the two currents I1 and I2. Once the total resistance is calculated, it can be used to solve for the individual currents.

## 3. Can I simplify the circuit to make it easier to solve?

Yes, it is possible to simplify the circuit by using series and parallel resistor combinations. This involves combining resistors that are in series (connected one after the other) or in parallel (connected across the same two nodes) to create a simpler circuit. This can make it easier to calculate the total resistance and solve for the currents.

## 4. How do I account for the internal resistance of the batteries?

The internal resistance of the batteries can be accounted for by including them as additional resistors in the circuit. The internal resistance will affect the total resistance in the circuit and therefore impact the calculated currents. It is important to consider the internal resistance when solving for the currents in a circuit with batteries.

## 5. What happens if I change the values of the batteries or resistors in the circuit?

Changing the values of the batteries or resistors in the circuit will affect the total resistance and therefore impact the calculated currents. It is important to recalculate the total resistance and solve for the currents whenever any changes are made to the circuit. Additionally, changing the values of the batteries may also impact the direction of the currents in the circuit.

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