Brooklyn College 1

DC Circuits- Kirchhoff’s Rules

Purpose

To study and apply Kirchhoff’s rules to ‘direct current’, ‘DC’ circuits.

Introduction

As we have learned in the previous experiments, the electric potential difference, or for brief the ‘voltage’, is related to

the ‘energy’ provided by the battery or voltage source. Due to conservation of energy, this voltage provided by the

battery distributes over the resistors in the circuit shown in fig. 1. The voltage provided by voltage source is called a

‘potential rise’, and the voltage across each of the resistors is called a ‘potential drop’. You might have guessed that due

to conservation of energy, the potential rise provided by the voltage source is equal to the sum of the potential drops

across the resistors. This is true and is not only true for the circuit of figure 1, but also for any closed loop in any circuit

with electrical components as resistors or also other electrical components. Kirchhoff’s Voltage law (or loop rule), ‘KVL’

which stems from the principle of conservation of energy, states that the potential rise provided by the sources is equal

to the sum of the potential drops across the components of any closed loop. Or the sum of all voltages across a closed

loop (taken with the correct polarity) is zero.

To apply Kirchhoff’s voltage rule, we label the points of the closed loop as shown in fig. 1. And label the currents

according to ‘branches’. A branch is any section of the loop that has all its components

in series, so they must have the same current (since current is the same for series

combination). In figure 1, the whole circuit and loop is one branch, with current equal to

the source current, 𝐼𝑠. Start with any point and label it, ‘a’. Choose the direction as clock

wise, ‘CW’ or Counter clock wise, ‘CCW’. If you encounter a component change the

letter after the component. If you encounter only a wire, then do not change the letter.

This is because the potential drop across a wire is very nearly equal to zero. Keep going

in the same direction until you return to the starting point.

Write the closed loop letters. For example ‘abcda’ then write the equation for KVL as

follows: If encounter a resistor and the current is in the same direction as your loop

direction, then take this voltage as (– 𝐼𝑅), otherwise take as (+𝐼𝑅). If you encounter a

battery, then take the polarity of the terminal of the battery that you end on. So, for our

loop in fig. 1, when we apply KVL starting from point a and going clock wise we have:

(−𝐼𝑠𝑅1) + (−𝐼𝑠𝑅2) + (−𝐼𝑠𝑅3) + (+𝑉𝑠) = 0 (1)

Kirchhoff has also a rule for currents. A ‘node’ is a point between different branches (see the definition of a circuit

‘branch’ above) in a circuit. The sum of currents entering a node, is equal to the sum of currents leaving the node. Stated

in another way, Kirchhoff’s current law (or junction rule), ‘KCL’ states that: The sum of all currents entering (or all

currents leaving) a node is equal to zero. The rule stems from the conservation of charge. The amount of charge that

flows into a note must continue to pass and flow out of the node.

Let’s apply both rules to the circuit of figure 2 below.

An important note about source currents is that the current that comes out of the battery (or any voltage source) must

also return back to the other terminal of the battery with the same value. So in figure 2, The current flowing from point

‘c’ to ‘e’ to the negative terminal of the battery 𝜀1 is 𝐼1. Also the current flowing from the wire ‘c’ to the negative

terminal of 𝜀2 is 𝐼2.

𝑅3

𝑅2

𝑅1

𝑉𝑠

a a

b

c d

𝐼𝑠

Figure 1: Applying

KVL for a simple loop

Brooklyn College 2

Figure 4: Analyzing a circuit using Kirchhoff’s

rules

First we label the branch currents. You can choose the directions of the currents in any way. If the calculation gives a

negative value of a current, then it means that this current is flowing in the opposite direction to the direction we

assumed. But the calculation and all the equations are still correct. The value of the current is correct and its direction is

opposite to the direction assumed. Label the points. Finally choose the closed loops and the direction either as CW or

CCW.

For the loop ‘acda’ we have KVL as:

−𝐼3𝑅3 + −𝐼1𝑅1 + +𝜀1 = 0 2

Rearranging we get: −𝐼1𝑅1 + 0 −𝐼3𝑅3 = −𝜀1 (2′)

For the loop ‘abca’ we have KVL as:

+𝐼2𝑅5 + −𝜀2 + +𝐼3𝑅3 = 0 (3)

Rearranging we get: 0 + 𝐼2𝑅5 + 𝐼3𝑅3 = 𝜀2 (3′)

Finally we need a third eqn. so we apply KCL at node ‘a’, we get: 𝐼1 + 𝐼2 + −𝐼3 = 0 (4′) 𝑜𝑟 𝐼1 + 𝐼2 = 𝐼3 (4)

If the values of the voltages of the batteries 𝜀1 and 𝜀2 and the values of the resistors are known then solving eqns. (2’),

(3’) and (4’) simultaneously we can calculate 𝐼1 , 𝐼2 & 𝐼3.

Current always flow from higher electric potential to lower electrical potential. When

we use a voltmeter, we need to connect its red terminal to the higher potential and its

black terminal to the lower potential. The voltage measured in fig. 3 is 𝑉𝑎𝑏 = 𝑉𝑎 − 𝑉𝑏 .

𝑉𝑎𝑏 is always equal to −𝑉𝑏𝑎 .

Running the experiment (The data sheet is on pages 5 & 6)

We are going to analyze the circuit of figure 4.

1) Open the simulator http://falstad.com/circuit/

Click ‘Run/ stop’ to stop the default simulation. Click ‘Circuits’ in

the top menu and select ‘Blank circuit’.

2) Click ‘Draw’ and select ‘Inputs and Sources’ and choose ‘Add

Voltage Source (2-terminal). Click and drag in the circuit space to

place it as in fig. 4. Move the mouse to the voltage source until its

color becomes light blue. Right click and edit its value to 5 (the unit

is ‘volts’ is already known by the simulator).

3) Click ‘Draw’ and select ‘Add Resistor’. Click and drag to add

resistors in the circuit (always while placing resistors, click and

drag from left to right for horizontal resistors, and from top to

bottom for vertical resistors. This is only a simulator note for

polarity but in actual lab the resistors do not have polarity). Keep adding resistors to build the circuit of fig. 4. Move the

mouse over each resistor till its color becomes light blue and right click and select ‘edit’ to set the values of each resistor

as shown in fig. 4, (do not type the unit, the simulator already has the unit). We will set the value of 𝑅𝑣𝑎𝑟 in step 8.

𝐼1 𝐼2

𝐼3

a

c d

a b

c

Figure 2: Applying Kirchhoff’s rules

Figure 3: Measuring

voltage using a voltmeter

𝑅𝑣𝑎𝑟

a

b

b c

𝜀

Brooklyn College 3

4) Click ‘Draw’ and select ‘Passive Components’ and choose ‘Add Switch’. Click and drag to place it as drawn in fig. 4.

Click it to open it.

5) Click ‘Draw’ and select ‘Add wire’. Connect wires between each two terminals of the components as shown in fig. 5.

The wire (and all components connections) has to connect to the square terminal of the component. Otherwise it will

give bad connection when we simulate. So from the left terminal of the 100 𝑜ℎ𝑚 you need 1 segment of wire to the left

terminal of the 10 𝑜ℎ𝑚, then another segment from the left terminal of the 10 𝑜ℎ𝑚 to the left terminal of the switch.

6) We now want to prepare to measure the voltages. Connect two wires across each component in the circuit, as shown

in fig. 6 below.

7) Click ‘Draw’, select ‘Outputs and labels’ then choose “Add Voltmeter/ Scope Probe’. Click and drag to connect the

voltmeters to the wires across each component, as shown in fig. 7. Make sure the polarity is correct as shown by the +

terminal of the voltmeter as in fig. 7. In a real circuit we would use a single voltmeter and move it to each component.

But in the simulator we can measure all the voltages simultaneously.

8) With the switch, S closed, (as shown in fig. 7, see the note in step 10 below), click ‘Run/Stop’ to simulate the circuit

then click it again to stop the simulation. The voltage values will still be displayed. Change the value of 𝑹𝒗𝒂𝒓, click ‘Run

/Stop’, read the voltage across 𝑅1 (the 100 𝑜ℎ𝑚) & repeat changing 𝑅𝑣𝑎𝑟 & running and stopping the simulator until the

Figure 5 (left): Connecting the circuit wires & Figure 6 (right): Preparing to measure voltages

Figure 7: The circuit

with voltmeters added

S a

b c

b c

d

e

Brooklyn College 4

voltage across the resistor 𝑅1 (the 100 𝑜ℎ𝑚) reads nearly 2v. You will find that this is achieved with 𝑅𝑣𝑎𝑟 = 54 𝑜ℎ𝑚𝑠.

Once this is adjusted, click ‘Run/Stop’ to simulate then click it again to stop the simulation. The voltage values will still be

displayed. Record the voltages values in the data sheet, in table 1. Divide each voltage by its corresponding resistor

value to find the currents. Record in table 1, in the data sheet. With S closed, the resistors, 𝑅2 10 𝛺 & 𝑅4 (22 𝛺) are in

series or in parallel?

9) Click ‘Run / Stop’ to stop the simulation, then click ‘Reset”. Now open the switch, S, and click ‘Run/ Stop’ to simulate

again. Read the voltages and record in table 2. Divide the voltages by the resistors to find the currents. Record in table 2.

With S opened, the resistors, 𝑅1 100 𝛺 & 𝑅2 10 𝛺 are in series or in parallel?

10) Now we will use KVL and KCL to calculate the currents. See the examples in the introduction part above.

With the switch S of fig. 4 closed, (note that point ‘e’ is the same as point ‘c’ when the switch, S is closed and can be

labeled ‘c’. We kept it as ‘e’ to make the circuit a general diagram for both cases of S closed and S opened), redraw the

circuit on page 6. First assign and draw the branch currents. Then Apply KVL for the closed loops. Finally apply KCL at

the nodes b and d.

Rearrange the equations and solve simultaneously for the values of the currents. Compare to the simulator values of

step 8.

Repeat for the case when the switch, S is opened. You will need to apply KCL for the node b. Compare with the

measured current values of step 9.

Questions

1. If the voltage source had a significant internal resistance, would this affect the results? Explain.

2. For the circuit of fig. 4, (with 𝑅𝑣𝑎𝑟 = 54 𝑜ℎ𝑚) with the switch, S opened, what is the total power dissipated in the

resistors? Show how you calculated this result (Hint: for questions 2 to 5, you can use your measured value of voltages,

𝑽’𝒔 and currents, 𝑰’𝒔 (= 𝑽/𝑹)).

3. For the circuit of fig. 4, (with 𝑅𝑣𝑎𝑟 = 54 𝑜ℎ𝑚) with the switch, S closed, what is the total power dissipated in the

resistors? Show how you calculated this result.

4. For the circuit of figure 4, (with 𝑅𝑣𝑎𝑟 = 54 𝑜ℎ𝑚) with the switch, S opened which resistor dissipates the most power?

Justify your answer.

5. For the circuit of figure 4, (with 𝑅𝑣𝑎𝑟 = 54 𝑜ℎ𝑚) with the switch, S closed which resistor dissipates the most power?

Justify your answer.

Brooklyn College 5

Data sheet

Name: Group: Date experiment performed:

Step 8) Table 1, measurements with switch S closed:

𝑉𝑎𝑏 (v) 𝐼𝑅𝑣𝑎𝑟 (𝐴)

𝑉𝑏𝑐 (v) 𝐼𝑅1 (𝐴)

𝑉𝑏𝑑 (v) 𝐼𝑅3 (𝐴)

𝑉𝑑𝑐 (v) 𝐼𝑅2 (𝐴)

𝑉𝑑𝑒 (v) 𝐼𝑅4 (𝐴)

With S closed, the resistors, 𝑅2 10 𝛺 & 𝑅4 (22 𝛺) are in series or in parallel?

Step 9) Table 2, measurements with switch S opened:

𝑉𝑎𝑏 (v) 𝐼𝑅𝑣𝑎𝑟 (𝐴)

𝑉𝑏𝑐 (v) 𝐼𝑅1 (𝐴)

𝑉𝑏𝑑 (v) 𝐼𝑅3 (𝐴)

𝑉𝑐𝑑 = −𝑉𝑑𝑐 (v) 𝐼𝑅2 (𝐴)

𝑉𝑑𝑒 (v) 𝐼𝑅4 (𝐴)

With S opened, the resistors, 𝑅1 100 𝛺 & 𝑅2 10 𝛺 are in series or in parallel?

Step 10) Use the space on the next page for the calculations of step 10.

For S closed, write the eqns. for KVL and KCL and solve for the currents, & compare with measured values:

For S opened, write the eqns. for KVL and KCL and solve for the currents, & compare with measured values:

Answers to questions

1.

2.

3.

4.

5. Continue on the next page

Brooklyn College 6

Use the space below for the equation and calculations of step 10)

With switch, S closed, redraw the circuit and draw the branch currents and label the points. Write & solve the eqns.:

With switch, S opened, redraw the circuit and draw the branch currents and label the points. Write & solve the eqns.: