homework_10_kirchoffs_rules
The circuit above contains 5 resistors [R_1, R_2, R_A, R_B, R_C] with equal resistance [R]. Write an equation for the current [i_2] in terms of the current [i_1]. ( when entering your answer, use i1 for [i_1], e.g., the answer 0.5*i1 has the correct syntax)
- [I_2 = \frac{ 1}{ 3} I_1]
- see octave code
Resistors [R_1, R_2, R_3, R_4] are arranged in a circuit as shown in the figure above. The direction of positive currents [I_1, I_2, I_3, I_4] through the resistors are shown. (i.e., negative current values correspond to currents in the opposite direction to that shown above). The values for the resistors and the batteries in the circuit are: [R_1 = 100.0 \Omega, R_2 = 200.0 \Omega, R_3 = 30.0 \Omega, R_4 = 400.0 \Omega, V_1 = 4.0 V, V_2 = 12.0 V]. What is the value of [I_2], the current through [R_2]? (remember [I_2] is a signed number)
[I_2 =]
The first step in solving physics problems is a CONCEPTUAL ANALYSIS. The first thing to consider is whether it is reasonable that there should be a current through [R_2]. Select all of the following statements that MUST be true in order to produce a current through [R_2] in the above circuit.
- There is at least one closed loop (completed circuit) containing R2.
- Check
- There is a battery in at least one closed loop containing R2.
- Check
- There is a battery in every closed loop containing R2.
- Wrong
Currents exist only in closed circuits and there must be some source of potential difference in that closed circuit. Here the battery is responsible for the potential difference. It is not necessary to provide a source in every closed circuit that contains [R_2], however. Consider for example a circuit consisting of a battery connected to the parallel combination of two resistors. There will be currents through both resistors but there will be one loop which has no battery (the loop containing just the two resistors).
Note that, in general, the source of potential difference would not have to be a battery. For example a circuit consisting of a resistor and an initially charged capacitor (which creates the potential difference) will have a current. This current will not be sustained, however, as the capacitor discharges and thereby the potential difference decreases in time. More on this possibility next week.
Now we know why it is reasonable to expect a current through R2 in this circuit. In order to calculate the value of this current, we will need to apply some laws of physics. Which one of the following laws of physics would you expect to be the most helpful to us in determining [I_2]?
- Kirchoff's Laws
Kirchhoff's Laws are designed exactly to make this kind of calculation. Kirchhoff's Laws should be the first thing you think of when confronted with a circuit problem. They always work. Of course, sometimes the problem is special enough that the circuit can be solved more naturally 'by inspection'.
Now we know that we will attempt to solve this problem using Kirchhoff's Laws. Before we leave this conceptual analysis section, however, let's be clear as to the source of Kirchhoff's Laws. They are not something 'new'. These Laws are really the application of more general laws to particular physical situations that involve electric circuits.
Kirchoff's Junction Law states that the sum of the currents into a junction is equal to the sum of the currents leaving the junction. This statement is a direct application to electric circuits of what more general physical principle?
- Conservation of Charge
Conservation of Charge states that there is no physical process that can create or destroy charge. When you 'create' a positive charge on a glass rod by rubbing it with a silk cloth, you also 'create' an equal amount of negative charge on the silk cloth.
If we apply this law to electric circuits we can say that charge cannot be created or destroyed at any point in the circuit. If we focus on a junction point, this law becomes Kirchhoff's Junction Law: the total current (or the charge in a specified amount of time) entering a junction is equal to the total current leaving the junction.
Now we know Kirchhoff's Junction Law is a direct application of the more general Principle of the Conservation of Charge. Kirchhoff's Loop Law states that the sum of the voltage drops around any closed loop is zero. This statement is a direct application to electric circuits of which of the following familiar concepts?
- The potential difference between two points is path independent
The power of the concept of Potential is that it is a 'property of the space'. A given distribution of charge produces an electric field that is defined at every point in space. The potential difference between any two points in space is determined by integrating that electric field between the two points. This integral is 'independent of the path' which is chosen between those two points.
We can directly apply this principle to electric circuits by first noting that the potential difference between a point and itself must be zero. Since the potential difference is independent of path, we can follow any closed loop in the circuit and simply add up the potential differences (voltage drops) across the circuit elements (resistors and batteries in this case) and we must get zero, the potential difference between a point and itself.
The next step in solving physics problems is a STRATEGIC ANALYSIS. We now need to develop a strategy for determining [I_2]. What is the minimum number of equations we need to write down to be able to apply Kirchhoff's Laws to this circuit to determine [I_2]?
- Two Loop Equations and One Junction Equation
At first glance it looks like we will need to apply Kirchhoff's Laws to three loops. However, on closer inspection, we see that the third loop ([V_2] and [R_4]) is not really coupled to the other loops. [R_3] couples the first two loops. Therefore we need to solve for three variables, [I_1], [I_2], and [I_3]. This procedure requires three equations. There is only one junction equation needed to relate [I_1], [I_2], and [I_3]. Consequently, we must find the other two equations from application of Kirchoff's Loop Law.
We have now completed the STRATEGIC ANALYSIS. We know we need to write down two loop equations and one junction equation in order to solve for [I_1], [I_2], and [I_3]. If you see how to do this, go ahead and choose your loops, write down the equations and solve them. If you would like some help with this procedure, click below to continue.
The final step in solving a physics problem is the QUANTITATIVE ANALYSIS Now we know we need to write three independent equations for [I1, I2, I3]. Let's begin with the one needed junction equation.
Which of the following equations can be obtained by an application of Kirchhoff's Junction Law to this circuit?
- [I_1 = I_2 + I_3]
If we apply Kirchoff's Junction Law to the point marked A below, we see that the total current coming into point A is [I_1] and the total current leaving point A is [I_2 + I_3]. Note that it may turn out that one or more of these currents will be found to be negative. For example, if [I_3] turns out to be negative, then it is really a current that is coming into point A rather than leaving it. That's just fine, however, a negative outgoing current is the same thing as a positive incoming current. It all comes out in the wash.
A circuit is constructed with six resistors and two batteries as shown. The battery voltages are [V_1 = 18.0 V] and [V_2 = 12.0 V]. The positive terminals are indicated with a [+] sign, The values for the resistors are: [R1 = R5 = 52.0 \Omega, R2 = R6 = 93.0 \Omega, R3 = 41.0 \Omega, R4 = 107.0 \Omega]. The positive directions for the currents [\I_1, I_2, I3] are indicated by the directions of the arrows.
A circuit is constructed with five resistors and one real battery as shown above right. We model. The real battery as an ideal emf [V = 12.0 V] in series with an internal resistance [r] as shown above left. The values for the resistors are: [R_1 = R_3 = 43.0 \Omega, R_4 = R_5 = 84.0 \Omega, R_2 = 150.0 \Omega]. The measured voltage across the terminals of the batery is [V_{battery} = 11.8 V].
