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Verification of OHM’s law By: Engr.Irshad Rahim Memon.

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Presentation on theme: "Verification of OHM’s law By: Engr.Irshad Rahim Memon."— Presentation transcript:

1 Verification of OHM’s law By: Engr.Irshad Rahim Memon

2  Objective of this practical is to verify that voltage applied to a closed resistive path is equal to the product of current flowing through and resistance of that path

3  Few resistors  bread board  multimeter  variable DC power supply  connecting wires etc

4  Ohm’s law states that current through a closed resistive path is directly proportional to applied voltage, where resistance of path is kept constant.  Ohm experimentally determined that if the voltage across resistor is increased, current through resistor will also increase as shown in figure 01.

5 Figure 01: Ohm law for constant resistance (a) less voltage, less current (b) more voltage, more current

6  Ohm also determined if voltage is kept constant then less resistance results in more current and more resistance results in less current as shown in figure 02

7  Figure 02: Ohm law for constant voltage (a) less resistance, more current (b) more resistance, less current

8  Ohm also determined if voltage is kept constant then less resistance results in more current and more resistance results in less current as shown in figure 02.

9  Ohm’s law can be formulated as follows:

10  Connect a variable voltage source, a resistor and an ammeter in series as shown in figure 03(a) to verify direct relationship between voltage and current as shown in figure 03(c)

11  Verification of Ohm’s law for constant resistance (a) circuit (b) different values of current for different values of voltage (c) graphical relation between voltage and current

12  Also connect a variable resistor, a voltage source and an ammeter in series as shown in figure 04(a) to verify inverse relationship between resistance and current as shown in figure 04(c)

13  It is observed that current varies directly with applied voltage where resistance is kept constant. It is also observed that current varies inversely with resistance where applied voltage is kept constant.

14  Fill the missing quantity in the table 01 with the help of Ohm’s law. S.No R (Ohm) I (Ampere) V (volt) Example1K22K 15M100 2 50 30.05 m1

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17 Verification of Kirchhoff’s Voltage Law (KVL) By: Engr.Irshad Rahim Memon

18  Objective of this practical is to verify KVL.

19  Few resistors  bread board  multimeter  variable DC power supply  connecting wires etc

20  Kirchhoff’s voltage law states that “sum of all the voltage drops around a single closed path in a circuit is equal to the total source voltage in that loop” or it can also be defined as “The algebraic sum of all the voltages (both source and drop) around a single closed path is equal to zero” as shown in figure 01.

21  Figure 01: Demonstration of KVL

22  KVL may be formulated as follows. Vs=V1+V2+V3+…+Vn or Vs-V1-V2-V3-…-Vn=0

23  For the verification of KVL, a simple series resistive circuit like shown in figure 01 should be prepared.

24  KVL is verified  KVL law is verified for series circuit of three resistors of known values connected in series. Results are given in table 01.

25 S.No Vs V R1 V R2 V R3 V s= V R1 +V R2 +V R3 example10Volt2 Volt3 Volt5 Volt10 Volt=2+3+5 Volt 15Volt 215volt 320volt

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27 Verification of Kirchhoff’s Current Law (KCL) By: Engr.Irshad Rahim Memon

28  Objective of this practical is to verify KCL.

29  Few resistors  bread board  multimeter  variable DC power supply  connecting wires etc

30  Kirchhoff’s current law states that “Algebraic sum of all currents entering and leaving a node is equal to zero” as demonstrated in figure 01.

31  Figure 01: Demonstration of KCL

32  For the verification of KCL, a simple parallel resistive circuit like shown in figure 02 should be prepared.

33  Figure 02: Verification of KCL

34  KCL law is verified for parallel circuit of three resistors of known values connected in parallel. Results are given in table 01.

35 S.No I total I R1 I R2 I R3 I total = I R1 +I R2 +I R3 example1Amp1Amp0.5 Amp0.3 Amp0.2 Amp1Amp=0.5+0.3+0.2 Amp 10.5 Amp 20.25 Amp0.25 Amp 30.1 Amp

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