Name of the Experiment: VERIFICATION OF KCL & CURRENT DEVIDER RULE
Course No. EEE 102 Experiment No. 3 Group No. 2
Mursalin Habib Roll No. 9906114 Department: Electrical and Electronics Engineering Level 1 Term 1 Session 1999-2000 Partners Roll No.
9906112 9906113 9906115 9906116
Date of Performance: Date of Submission:
Bangladesh University of Engineering and Technology
OBJECTIVE:
The experiment is intended to verify Kirchhoff’s current law (KCL) with the help of parallel circuits and hence derive its equivalent resistance both experimentally and analytically.
THEORY: If a circuit has a number of interconnected branches, two other laws are applied in order to find the current flowing in the various branches. These laws, discovered by the German physicist Gustav Robert Kirchhoff, are known as Kirchhoff's laws of networks. The first of Kirchhoff's laws which is known as KCL states that at any junction in a circuit through which a steady current is flowing, the sum of the currents flowing to the point is equal to the sum of the currents flowing away from that point. The second law states is known as KVL.
Simply, KCL states the algebraic sum of currents entering any node equals the sum of the currents leaving the node.
∑
Ientering =
∑
Ileaving
The current divider rule is given by I X
=
R P R X
×
I S
The equivalent (total) resistance of a parallel circuit is given by the formula. 1 R p
=
1 R1
+
1 R 2
+
1 R3
APPARATUS: • • • • • •
One DC Voltmeter(0—300V) One DC Ammeter(0—5A) Two lamp boards(220V—100W) work as rheostat Four SPST switches One multimeter DC power supply
I0 I + -
I1
S2
S3
I2
V
I1
I2
Rp PROCEDURE:
1. Two rheostats along with there own switches ware connected in parallel across the power supply as shown in the figure. 2. 30V DC applied from DC power supply. 3. KCL (I=I1+I2+I3) was verified for each set of data. And total resistance of the parallel circuit was calculated. . And it was compared with experimentally obtained value Rp = Vs /I. It was also verified with all the set of data.
TABLE:
Obs. No 1 2 3 4 5
Vs Volt 174 175.5 175.5 175.5 177
I Amp 0.38 0.615 0.85 0.99 1.155
CALCULATION:
For First observation: (Sample) R0 = Vs /(I1+I2+I3) ⇒ R0 = 174/(0.231+0.1545) ⇒ RO= 451.36 Ω
I1 Amp 0.231 0.46 0.695 0.695 0.695
I2 Amp 0.1545 0.15 0.15 0.305 0.46
I1+I2 Amp 0.3855 0.61 0.854 1.0 1.155
RP Ohms 451.36 287.70 207.69 175.5 153.25
Req=V0 /I Ohms 457.89 285.37 206.47 177.27 153.25
REPORT:
Comment on the result obtained and discrepancies. (if
any)
Results obtained: For each observation, the current, ‘I’ is obtained by placing an ammeter in a series combination with the whole circuit. V 0 is obtained by the indicator of the constant power source. Confirmation was made of V o by measuring the voltage of the power source with a voltmeter. The summation of the resistance of the rheostats was R o.
And the division of V s by I was Rp. Discrepancies: There are two reasons for discrepancies. Reading error: While a meter obtains the reading, a least error of obtaining reading can’t be omitted. So whenever we obtained a reading by a meter we get a least error. This error is summed up after any calculation that was performed with these readings. As the least error of reading of a multi meter is different from the least error of the ammeter or voltmeter, which is also different. So whenever we divide the V o by I we get a little difference result from summating the individual resistance of the rheostats. Moreover as we mean the connecting wires has no resistance, practically is not true; so whenever we try to measure the total current it differs from the summation of the current passing through in every resistance connecting in parallel.
Loading effect of the meter: A voltmeter has an infinite resistance and an ammeter has a resistance of zero – is theoretically true but not in practical use. Practically a voltmeter has a very big resistance and an ammeter has a very small one. So whenever an ammeter is placed in the circuit, the equivalent resistance differs slightly & the total current of the circuit with an ammeter also differs for this. Again whenever a voltmeter is placed to obtained the voltage the voltage differs from the voltage of the rheostats without a voltmeter connected in parallel.
QUESTION & ANSWER:
1. Show analytically that for a parallel circuit, 1 R p
=
1 R1
+
1 R 2
+
1 R3
According to figure, I=I1+I2+I3 According to ohms law, I1R1=I2R2= I3R3=Vs
So, 1
I=
+
R1
I
=
V s
I R
1 R 2
=
+
1
Vs
R3
+
R1
1 1
R 2
+
1 R3
so, 1 R p
=
1
+
R1
1 R 2
+
1 R3
is proved.
2. “KCL is a restatement of the law of conservation of charge”-Justify the statement
We know that, P = VI
⇒P=
VQ t
As the circuit is arranged in parallel, So, P=P1+P2+P3
⇒
VQ
=
VQ1
+
t t ⇒ Q = Q + Q2
VQ2 t + Q3
+
VQ3 t
“KCL is a restatement of the law of conservation of charge”-Justified. 3. “KCL is applicable in closed surface too”—Explain Let us imagine abcd is a closed surface. Where there is an electrical circuit on it. Current I1,I2,I3 is entering the surface through a,b,c, & and I 4 is leaving through d .let us think that abcd is a node then we can say according to KCL -i1-i2-i3+i4=0
So, KCL is applicable in a closed surface too.
Discussion:
To analyze a complex electric circuits KCL is a must. Because it bears implication for analyzing a particular node of a complex circuit. So, it is really important to learn how to apply KCL to a simple circuit. So, the verification of KC L make us a bit skillful to design a complex circuit. In the practical field, the idea of the results of the discrepancies makes us able to be alert of designing an electric circuit. An it was really encouraging for a freshman student like me to use those sophisticated Electrical Elements.
