# Kirchhoff's Law

**Topics:**Ohm's law, Electromotive force, Electric current

**Pages:**5 (1024 words)

**Published:**March 13, 2013

II. Objectives: To study the application of Kirchhoff’s Law to a D.C. network by comparing the

observed and the computed values of the currents in the circuit. III. Apparatus: Resistance Module, 1pc. Battery of two cells (3 volts), 1pc. Dry cell (1.5 volts),

Multitester, 4 pairs connectors

IV. Procedure with Experimental Setup:

Part A.

1.) The apparatus was arranged as in diagram 1. The switches were left open until it is checked. 2.) With the switches closed the voltages V1 and V2 across the batteries were measured with a voltmeter. 3.) Considering V1, V2, R1, R2 and R3 as known, the Kirchhoff’s law was applied to the circuits and the correct number of independent current and voltage equations necessary to solve for the unknown current in each branch was written, that is one current equation and the fewest number of voltage equations that will include every emf and every resistance at least once in a set of equations. The equations for these currents were solved. 4.) Having calculated the current in each brach by an application of Kirchhoff’s Laws, the respective currents were measured experimentally. The circuit was break in turn in each branch and the ammeter was inserted in series to observe the values of the currents. 5.) The percentage error between the observed values of the currents and those computed by Kirchhoff’s Laws were calculated. Part B.

1.) The series of observation and calculations were repeated using the arrangement in figure 2. V. Tabulated Data and Results

Part A.

R1=2.6 Ω

R2=3.0 Ω

R3=3.3 Ω

V1=1.373 V

V2=2.648 V

Resistors| I (comp.)| I (obs.)| Deviation| % Error|

R1| 3.368 mA| 3.4 mA| 99.6| 1|

R2| 422.081 mA| 425 mA| 49.996| 0.7|

R3| 418.714 mA| 418.6 mA| 50.395| 0.3|

Part B.

R4=11.2 Ω

V1=1.373 V

V2=2.648 V

Resistors| I (comp.)| I (obs.)| Deviation| % Error|

R1| 0.263| 0.27| 70.968| 3|

R2| 0.23| 0.25| 73.118| 8|

R3| 0.209| 0.22| 76.344| 5|

R4| 0.175| 0.19| 79.56| 8|

VI. Computations

R1=2.6 Ω

R2=3.0 Ω

R3=3.3 Ω

R4=11.2 Ω

V1=1.373 V

V2=2.648 V

Part A.

1.) Solving for currents I1, I2, and I3

* Kirchhoff’s first rule (Junction Rule):

I1=I2+I3

I1-I2-I3=0 Eq. 1

* Kirchhoff’s second rule (Loop Rule):

Upper loop (Counterclockwise)

I1R1+I2R2-Va-Vb=0

I1R1+I2R2=Va-Vb Eq. 2

Lower loop (Counterclockwise)

I3R3-I2R2-Vb=0

I3R3-I2R2=Vb Eq. 3

by Substitution and Elimanation Method,…

I1=-3.368 mA so I1 is not CCW but CW

I2=-422.081 mA so I2 is not ->but<-

I3=418.714 mA since I3 is positive it is CCW

Part B.

1.) Solving for currents I1, I2, I3, and I4

* Kirchhoff’s first rule (Junction Rule):

I1=I2+I3+I4

I1-I2-I3-I4=0 Eq. 1

* Kirchhoff’s second rule (Loop Rule):

I1R1+I2R2=Va Eq. 2

I1R1+I3R3=Va Eq. 3

I1R1+I4R4=Va-Vb Eq. 4

then,..

I1=Va-I1R1R2+Va-I1R1R3+Va-Vb-I1R1R4

by Substitution and Elimanation Method,…

I1=0.263 A since I3 is positive its directions is correct

I2=0.23 A since I3 is positive its directions is correct

I3=0.209 A since I3 is positive its directions is correct

I4=0.175 A since I4 is negative its directions is opposite

VII. Questions and Problems

1.) What is the effect caused by the ammeter resistance when an ammeter is inserted into a circuit to measure the current? An ammeter has a finite resistance which is inserted in series with the rest of the circuit, increasing the total resistance and decreasing the current but since the effective resistance of ammeter is so small that it can be neglected and the current in the circuit is almost the same as it was when the ammeter was not connected. 2.) Two cells of emf 1.5 and 2.0 volts and internal resistance 0.5 and 0.8 ohm respectively are connected in parallel. Determine what current they will deliver to external 2.0 ohms, use the Kirchhoff’s Laws. * Kirchhoff’s first rule (Junction Rule):

I1+I2+I3=0...

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