Series And Parallel Circuits.docx

SERIES AND PARALLEL CIRCUITS Muhammad Yusuf, Asraeni, Eti Puspasari. Laboratory of Fundamental Physics Department of Physics FMIPA Makassar State University Abstract Have done an experiments "Series and Parallel circuit" with the aim to skilled in arranging resistror become in series and parallel arrangement, to place and use basicmeter with properly, and to understand the principles of the laws kirchoof and understand the characteristics of series and parallel circuits resistor. The tools used in this experiment is the power supply AC/DC 0-12 V, resistors with different values, basicmeter and connecting wire. In this experiment there are two activities. The first activity is the resistor series circuit, in these activities are arranged circuit in series resistor that connects the two resistors with a power supply with uses a connecting cable and a strong measured current and voltage in the circuit by using basicmeter. The second activity is a series of parallel resistor, the activity is done the same with the first activity, it's just a different circuit is arranged in parallel resistor. From this experiment, the equivalent resistance for the resistor series arrangement is Req = R1 + R2 + . . . while the equivalent resistance of the parallel 1 1 1 1 resistor arrangement is = + + + . . . . Thus, it can be concluded that the function of the 𝑅𝑒𝑞

𝑅1

𝑅2

𝑅3

resistor in series arrangement is a voltage divider resistors and function with parallel arrangement is current divider.

KEYWORD : Resistor, rangkaian, tegangan, kuat arus, resistansi. PROBLEMS STATEMENT 1. How to string up a resistor become series and parallel arrangement ? 2. How to use basicmeter correctly ? 3. How is the principle of the laws of kirchoff ? 4. How is the characteristics of the resistor of series and parallel circuits ? OBJECTIVES 1. Students skilled in string up resistor become series and parallel arrangement. 2. Students can put and use basicmeter with correctly. 3. Students can understand the principles of the laws of kirchoff. 4. Students can understand the characteristics of series and parallel resistor circuit. EXPERIMENTS METHODOLOGY Theory Two or more resistors connected in such a way that the same charge must be coursing through both said that the resistors connected in series. Equivalent resistance are given a set of resistor is equal to the sum of the resistance's, ie;

Req = R1 + R2 + R3 + . . . .

(1.1)

(Tipler, 2001:154). Two resistors connected in such a way so as to have the same voltage between the two is said that the resistors are connected in parallel. Equivalent resistance is equal to the sum of the inverse of resistance each, namely; 1 𝑅𝑒𝑞

1

1

1

=𝑅 +𝑅 +𝑅 +.... 1

2

3

(1.2)

(Tipler, 2001: 154-155). Results of potential difference measurements on resistors R1 and R2 (different values) are arranged in series show different results, however, if the measured current through the resistor is then obtained by the same measurement. By contrast, if the resistors are arranged in parallel, different measurement results obtained. The current through each resistor is different, but the measurement of the voltage across each resistor is the same. This fact shows that this type of arrangement determine the value of the variable resistor voltage and strong electrical current in the circuit. In the arrangement of the series, a resistor serves as a voltage divider, which means if the voltage across each resistor dijumlhkan then the amount is equal to the magnitude of the voltage source. Meanwhile, if the resistors are arranged in parallel, it serves as a divider resistor current, which means that if a strong electric current passing through each resistor is measured, then it will have the same value as the total current before branching point (Herman, 2015: 21). According to Abdul Haris Bakri et al (2008: 25-26), there are two laws that apply to the circuit which has a fixed flow. Where the law is better known by the name of Kirchoff's law, namely: 1. At each closed circuit, the algebraic sum of the potential differences must be equal to zero. 2. At each branching point, the amount of current that goes through the point is equal to the amount of current that is out of the point. The first Kirchoff law of voltage loop is often called the law, because in fact the potential difference between two points in a circuit in the steady state is

always constant. Kirchoff's second law of currents known as the legal ramifications, because it meets the law of conservation of charge. This law is necessary for the circuit multisimpal containing branching point when the current begins to split. At steady state, there is no accumulation of electric charge at any point in the circuit, thus the number of incoming cargo at some point will leave that point by the same amount (Abdul Haris Bakri, 2008: 26-28). Equipment List A. Tools 1. 1 piece of power supply AC/DC, 0 – 12 V 2. 2 pieces of resistors with different values. 3. 1 pieces of basicmeter. 4. Connector wire. B. Material Identification of Variables Activity 1. Series resistors circuit. a. The first resistor / R1 (ohm) b. The second resistor / R2 (ohm) c. Source voltage (V) d. Electric current before R1 (A) e. Electric current between R1 and R2 (A) f. Electric current after R2 (A) g. The voltage on R1 (V) h. The voltage on R2 (V) Activity 2. Parallel resistor circuit. a. The first resistor / R1 (ohm) b. The second resistor / R2 (ohm) c. Source voltage (V) d. Electric current total (before the branch point) (A) e. Electric current through R1 (A) f. Electric current through R2 (A)

g. The voltage on R1 (V) h. The voltage on R2 (V) Operational Definition of Variables Activity 1. Series resistors circuit. a. The first resistor / R1 is a device mounted on the circuit labeled first resistor that serves as a barrier to the flow of the unit ohm. b. The second resistor / R2 is a device mounted on the circuit labeled second resistor that serves as a barrier to the flow of the unit ohm. c. Source voltage is the voltage used in the power supply as a voltage divider in the circuit which uses the unit Volt. d. Electric current before R1 is the amount of current flowing in the circuit before passing the first resistor is measured using basicmeter in units of Ampere. e. Electric current between R1 and R2 is the amount of current flowing in the circuit between the first resistor and second resistor are measured using basicmeter in units of Ampere. f. Electric current after R2 is the amount of current flowing in the circuit after passing through the second resistor is measured using basicmeter in units of Ampere. g. The voltage on R1 is the magnitude of the voltage at the end of the first resistor is measured using basicmeter in units of Volt. h. The voltage on R2 is the magnitude of the voltage at the end of the second resistor is measured using basicmeter in units of Volt. Activity 2. Parallel resistor circuit. a. The first resistor / R1 is a device mounted on the circuit labeled first resistor that serves as a barrier to the flow of the unit ohm. b. The second resistor / R2 is a device mounted on the circuit labeled second resistor that serves as a barrier to the flow of the unit ohm. c. Source voltage is the voltage used in the power supply as a voltage divider in the circuit which uses the unit Volt.

d. Electric current total (before the branch point) is the amount of current flowing in the circuit before the branch point is measured using basicmeter in units of Ampere. e. Electric current through R1 is the amount of current flowing in the circuit that passes through the first resistor and measured using basicmeter in units of Ampere. f. Electric current through R2 is the amount of current flowing in the circuit that passes through the second resistor and measured using basicmeter in units of Ampere. g. The voltage on R1 is the magnitude of the voltage at the end of the first resistor is measured using basicmeter in units of Volt. h. The voltage on R2 is the magnitude of the voltage at the end of the second resistor is measured using basicmeter in units of Volt. Procedure Activity 1. Series resistors circuit. 1. The experimental device coupled in series with a second resistor connected to the power supply and basicmeter are using a connecting cable. 2. Strong current (before the first resistor, between the first and the second resistor, and after the second resistor) is measured using basicmeter. The results are recorded in table observations. 3. The voltage across the first resistor and the second resistor is measured using basicmeter. The results are recorded in table observations. 4. The move is performed by a different source voltage value and the result recorded in the observations. Activity 2. Parallel resistor circuit. 1. The experimental device are arranged in parallel with two resistors connected to the power supply and basicmeter are using a connecting cable. 2. Strong current (before the branching point, through the first resistor, and through the second resistor) is measured using basicmeter. The results are recorded in table observations.

3. The voltage across the first resistor and the second resistor is measured using basicmeter. The results are recorded in table observations. 4. The move is performed by a different source voltage value and the result recorded in the observations. RESULT OF EXPERIMENT AND DATA ANALYSIS Result of Experiment Activity 1. Series resistors circuit. R1

R2

V A

Figure 1.1. Series resistors circuit. R1 = 100 ohm. R2 = 150 ohm. Table 1. Results of measurements of electric current and voltage at a resistor series circuit. Source Electric Current (A) Voltage on Voltage on Voltage Between R1 & R2 R1 (V) R2 (V) Before R1 After R2 (V) 1. │3,0 ± 0,1│ │0,016 ± 0,001│ │0,016 ± 0,001│ │0,016 ± 0,001│ │1,2 ± 0,1│ │1,8 ± 0,1│ 2. │6,0 ± 0,1│ │0,032 ± 0,001│ │0,032 ± 0,001│ │0,032 ± 0,001│ │2,4 ± 0,1│ │3,6 ± 0,1│ 3. │9,0 ± 0,1│ │0,048 ± 0,001│ │0,048 ± 0,001│ │0,048 ± 0,001│ │3,6 ± 0,1│ │5,4 ± 0,1│ Activity 2. Parallel resistor circuit.

No .

R1 = 100 ohm R2 = 150 ohm Table 1. Results of measurements of electric current and voltage at a resistor parallel circuit. No . 1.

Electric Current (A) Source Voltage Total (before the Through R1 Through R2 (V) branch point) │3,0 ± 0,1│ │0,04 ± 0,01│ │0,02 ± 0,01│ │0,02 ± 0,01│

Voltage on R1 (V)

Voltage on R2 (V)

│3,0 ± 0,1│ │3,0 ± 0,1│

2. 3.

│6,0 ± 0,1│ │0,08 ± 0,01│ │0,05 ± 0,01│ │9,0 ± 0,1│ │0,14 ± 0,01│ │0,08 ± 0,01│ Analysis Data Activity 1. Series resistors circuit. R1

│0,04 ± 0,01│ │0,06 ± 0,01│

R2

V A

Rtotal = R1 + R2 Rtotal = 100 ohm + 150 ohm Rtotal = 250 ohm 1. Electric current. a. Electric current with voltage │3,0 ± 0,1│V. I1 = 𝑅

𝑉1

3,0 V

𝑡𝑜𝑡𝑎𝑙

= 250 ohm = 0,012 A

So that, I1 = 𝑅

𝑉1 𝑡𝑜𝑡𝑎𝑙

I1 = V1 × Rtotal-1 𝜕𝐼

𝜕𝐼1

dI1 = |𝜕𝑉1 d𝑉1 | + |𝜕𝑅 1

𝑡𝑜𝑡𝑎𝑙

d𝑅𝑡𝑜𝑡𝑎𝑙 |

dI1 = |𝑅𝑡𝑜𝑡𝑎𝑙 −1 d𝑉1 | + |𝑉1 × 𝑅𝑡𝑜𝑡𝑎𝑙 −2 d𝑅𝑡𝑜𝑡𝑎𝑙 | d𝐼1 𝐼1 d𝐼1 𝐼1 ∆𝐼1 𝐼1

𝑅𝑡𝑜𝑡𝑎𝑙 −1

=|

𝐼1

d𝑉

𝑉1 ×𝑅𝑡𝑜𝑡𝑎𝑙 −2

d𝑉1 | + |

𝐼1

d𝑅

= | 𝑉1 | + | 𝑅 𝑡𝑜𝑡𝑎𝑙 | 1

∆𝑉

𝑡𝑜𝑡𝑎𝑙

= | 𝑉1 +

∆𝑅𝑡𝑜𝑡𝑎𝑙

∆𝑉

∆𝑅𝑡𝑜𝑡𝑎𝑙

1

∆𝐼1= | 𝑉1 + 1

𝑅𝑡𝑜𝑡𝑎𝑙

𝑅𝑡𝑜𝑡𝑎𝑙

| | 𝐼1

Because Rtotal is constant, so that;

d𝑅𝑡𝑜𝑡𝑎𝑙 |

│6,0 ± 0,1│ │6,0 ± 0,1│ │9,0 ± 0,1│ │9,0 ± 0,1│

∆𝑉

∆𝐼1= | 𝑉1 | 𝐼1 1

0,1 V

∆𝐼1= | 3,0 V | 0,012 A ∆𝐼1= 0,033 × 0,012 A ∆𝐼1= 0,0004 A RE =

∆𝐼1

× 100 % =

𝐼1

0,0004 A 0,012 A

× 100 % = 3,33 % (3 Significant Figures)

Reporting of physics: PR = │𝐼1 ± ∆𝐼1│ 𝐼1 = │0,012000 ± 0,000400│A 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,012000 A − 0,016 A

=|

0,014 A

= |−

| × 100 %

| × 100 %

0,004 A

| × 100 %

0,014 A

= 28,57 % b. Electric current with voltage │6,0 ± 0,1│V. I2 = 𝑅

𝑉2

6,0 V

𝑡𝑜𝑡𝑎𝑙

= 250 ohm = 0,024 A

So that, ∆𝑉

∆𝐼2 = | 𝑉2 | 𝐼2 2

0,1 V

∆𝐼2 = | 6,0 V | 0,024 A ∆𝐼2 = 0,017 × 0,024 A ∆𝐼2 = 0,0004 A RE =

∆𝐼2 𝐼2

× 100 % =

0,0004 A 0,024 A

Reporting of Physics: PR = │𝐼2 ± ∆𝐼2 │

× 100 % = 1,67 % (3 Significant Figures)

𝐼2 = │0,024000 ± 0,000400│A 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,024000 A− 0,032 A

=|

0,028 A

= |−

| × 100 %

| × 100 %

0,008 A

| × 100 %

0,028 A

= 28,57 % c. Electric current with voltage │6,0 ± 0,1│V. I3 = 𝑅

𝑉3

9,0 V

= 250 ohm = 0,036 A

𝑡𝑜𝑡𝑎𝑙

So that, ∆𝑉

∆𝐼3 = | 𝑉3 | 𝐼3 3

0,1 V

∆𝐼3 = | 9,0 V | 0,036 A ∆𝐼3 = 0,011 × 0,036 A ∆𝐼3 = 0,0004 A RE =

∆𝐼3

× 100 % =

𝐼3

0,0004 A 0,036 A

× 100 % = 1,11 % (3 Significant Figures)

Reporting of Physics: PR = │𝐼3 ± ∆𝐼3 │ 𝐼3 = │0,036000 ± 0,000400│A 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,036000 A− 0,048 A

=|

= |−

0,042 A 0,012 A

| × 100 %

| × 100 %

0,042 A

= 28,57 %

| × 100 %

2. Voltage. a. Voltage with electric current │0,016 ± 0,001│A V1.1 = 𝐼1 × 𝑅1 = 0,016 A × 100 ohm = 1,6 V So that, V1.1 = 𝐼1 × 𝑅1 𝜕𝑉

𝜕𝑉

dV1.1 = | 𝜕𝐼1.1 d𝐼1 | + | 𝜕𝑅1.1 d𝑅1 | 1

1

dV1.1 = |𝑅1 d𝐼1 | + | 𝐼1 d𝑅1 | d𝑉1.1 𝑉1.1 d𝑉1.1 𝑉1.1 ∆𝑉1.1 𝑉1.1

𝑅

𝐼

= |𝑉 1 d𝐼1 | + |𝑉 1 d𝑅1 | 1.1

=|

d𝐼1 𝐼1

1.1

d𝑅

|+|𝑅1| 1

= | 𝐼1 +

∆𝐼

∆𝑅1

∆𝐼

∆𝑅1

1

∆𝑉1.1= | 𝐼 1 + 1

𝑅1

𝑅1

| | 𝑉1.1

Because R1 is constant, so that; ∆𝐼

∆𝑉1.1= | 𝐼 1 | 𝑉1.1 1

0,001 A

∆𝑉1.1= | 0,016 𝐴 | 1,6 V ∆𝑉1.1= 0,0625 × 1,6 V ∆𝑉1.1= 0,1 V RE =

∆𝑉1.1 𝑉1.1

× 100 % =

0,1 𝑉 1,6 V

× 100 % = 6,25 % (2 Significant Figures)

Reporting of Physics: PR = │𝑉1.1 ± ∆𝑉1.1│ 𝑉1.1 = │1,60 ± 0,10│V 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

1,60 V− 1,2 V

=|

1,4 V

| × 100 %

| × 100 %

0,4 V

= |1,4 𝑉| × 100 % = 28,57 % V1.2 = 𝐼1 × 𝑅2 = 0,016 A × 150 ohm = 2,4 V So that, ∆𝐼

∆𝑉1.2= | 𝐼 1 | 𝑉1.2 1

0,001 A

∆𝑉1.2= | 0,016 𝐴 | 2,4 V ∆𝑉1.2= 0,0625 × 2,4 V ∆𝑉1.2= 0,15 V RE =

∆𝑉1.2 𝑉1.2

× 100 % =

0,15 𝑉 2,4 V

× 100 % = 6,25 % (2 Signifcant Figures)

Reporting of Physics: PR = │𝑉1.2 ± ∆𝑉1.2│ 𝑉1.2 = │2,40 ± 0,15│V 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

2,40 V− 1,8 V

=|

1,4 V

| × 100 %

| × 100 %

0,6 V

= |2,1 𝑉| × 100 % = 28,57 % Thus, the total voltage on the resistor is Vtotal.1 = V1.1 + V1.2 Vtotal.1 = 1,60 V + 2,40 V Vtotal.1 = 4,00 V So that, Vtotal.1 = V1.1 + V1.2

𝜕𝑉𝑡𝑜𝑡𝑎𝑙.1

dVtotal.1 = |

𝜕𝑉1.2

𝜕𝑉𝑡𝑜𝑡𝑎𝑙.1

d𝑉1.1 | + |

𝜕𝑉1.2

d𝑉1.2 |

dVtotal.1 = |1 d𝑉1.1 | + |1 d𝑉1.2 | ∆Vtotal.1 = |∆𝑉1.1 | + |∆𝑉1.2 | ∆Vtotal.1 = 0,10 V + 0,15 V ∆Vtotal.1 = 0,25 V RE =

∆𝑉𝑡𝑜𝑡𝑎𝑙.1 𝑉𝑡𝑜𝑡𝑎𝑙.1

0,25 𝑉

× 100 % = 4,00 V × 100 % = 6,25 % (2 Significant Figures)

Reporting of Physics: PR = │𝑉𝑡𝑜𝑡𝑎𝑙.1 ± ∆Vtotal.1│ 𝑉𝑡𝑜𝑡𝑎𝑙.1 = │4,00 ± 0,25│V b. Voltage with electric current │0,032 ± 0,001│A V2.1 = 𝐼2 × 𝑅1 = 0,032 A × 100 ohm = 3,2 V So that, ∆𝐼

∆𝑉2.1= | 𝐼 2 | 𝑉2.1 2

0,001 A

∆𝑉2.1= | 0,032 𝐴 | 3,2 V ∆𝑉2.1= 0,03125 × 3,2 V ∆𝑉2.1= 0,1 V RE =

∆𝑉2.1 𝑉2.1

× 100 % =

0,1 𝑉 3,2 V

× 100 % = 3,125 % (3 Significant Figures)

Reporting of Physics: PR = │𝑉2.1 ± ∆𝑉2.1│ 𝑉2.1 = │3,200 ± 0,100│V 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

3,200 V− 2,4 V

=|

1,4 V

| × 100 %

| × 100 %

0,8 V

= |2,8 𝑉| × 100 % = 28,57 % V2.2 = 𝐼2 × 𝑅2 = 0,032 A × 150 ohm = 4,8 V So that, ∆𝐼

∆𝑉2.2= | 𝐼 2 | 𝑉2.2 2

0,001 A

∆𝑉2.2= | 0,032 𝐴 | 4,8 V ∆𝑉2.2= 0,03125 × 4,8 V ∆𝑉2.2= 0,15 V RE =

∆𝑉2.2 𝑉2.2

× 100 % =

0,15 𝑉 4,8 V

× 100 % = 3,125 % (3 Significant Figures)

Reporting of Physics: PR = │𝑉2.2 ± ∆𝑉2.2│ 𝑉2.2 = │4,800 ± 0,150│V 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

| × 100 %

4,8 V− 3,6 V

=|

4,2 V

| × 100 %

1,2 V

= |4,2 𝑉| × 100 % = 28,57 % Thus, the total voltage on the resistor is Vtotal.2 = V2.1 + V2.2 Vtotal.2 = 3,200 V + 4,800 V Vtotal.2 = 8,000 V So that, ∆Vtotal.2 = |∆𝑉2.1 | + |∆𝑉2.2 |

∆Vtotal.2 = 0,100 V + 0,150 V ∆Vtotal.2 = 0,250 V RE =

∆𝑉𝑡𝑜𝑡𝑎𝑙.2 𝑉𝑡𝑜𝑡𝑎𝑙.2

0,250 𝑉

× 100 % = 8,000 V × 100 % = 3,13 % (3 Significant Figures)

Reporting of Physics: PR = │𝑉𝑡𝑜𝑡𝑎𝑙.2 ± ∆Vtotal.2│ 𝑉𝑡𝑜𝑡𝑎𝑙.2 = │8,000 ± 0,250│V c. Voltage with electric current │0,048 ± 0,001│A V3.1 = 𝐼3 × 𝑅1 = 0,048 A × 100 ohm = 4,8 V So that, ∆𝐼

∆𝑉3.1= | 𝐼 3 | 𝑉3.1 3

0,001 A

∆𝑉3.1= | 0,048 𝐴 | 4,8 V ∆𝑉3.1= 0,021 × 4,8 V ∆𝑉3.1= 0,099 V RE =

∆𝑉3.1 𝑉3.1

0,099 𝑉

× 100 % =

4,8 V

× 100 % = 2,1 % (3 Significant Figures)

Reporting of Physics: PR = │𝑉3.1 ± ∆𝑉3.1│ 𝑉3.1 = │4,8000 ± 0,0990│V 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

| × 100 %

4,8000 V− 3,6 V

=|

1,4 V

| × 100 %

1,2 V

= |4,2 𝑉| × 100 % = 28,57 % V3.2 = 𝐼3 × 𝑅2 = 0,048 A × 150 ohm = 7,2 V

So that, ∆𝐼

∆𝑉3.2= | 𝐼 3 | 𝑉3.2 3

0,001 A

∆𝑉3.2= | 0,048 𝐴 | 7,2 V ∆𝑉3.2= 0,02083 × 7,2 V ∆𝑉3.2= 0,1499 V RE =

∆𝑉3.2 𝑉3.2

× 100 % =

0,1499 𝑉 7,2 V

× 100 % = 2,1 % (3 Significant Figures)

Reporting of Physics: PR = │𝑉3.2 ± ∆𝑉3.2│ 𝑉3.2 = │7,200 ± 0,149│V 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

7,200 V− 5,4 V

=|

6,3 V

| × 100 %

| × 100 %

1,8 V

= |6,3 𝑉| × 100 % = 28,57 % Thus, the total voltage on the resistor is Vtotal.3 = V3.1 + V3.2 Vtotal.3 = 4,800 V + 7,200 V Vtotal.3 = 12,000 V So that, ∆Vtotal.3 = |∆𝑉3.1 | + |∆𝑉3.2 | ∆Vtotal.3 = 0,0990 V + 0,149 V ∆Vtotal.3 = 0,248 V RE =

∆𝑉𝑡𝑜𝑡𝑎𝑙.3 𝑉𝑡𝑜𝑡𝑎𝑙.3

0,248 𝑉

× 100 % = 12,000 V × 100 % = 2,07 % (3 Significant Figures)

Reporting of Physics: PR = │𝑉𝑡𝑜𝑡𝑎𝑙.3 ± ∆Vtotal.3│ 𝑉𝑡𝑜𝑡𝑎𝑙.3 = │12,000 ± 0,248│V Thus, it can be knowed that; 𝑉𝑠 = 𝑉1 + 𝑉2 𝐼𝑡𝑜𝑡 = 𝐼1 = 𝐼1 𝑉 = 𝐼𝑅 𝐼𝑡𝑜𝑡 𝑅𝑡𝑜𝑡 = 𝐼1 𝑅1 + 𝐼2 𝑅2 𝑅𝑡𝑜𝑡 = 𝑅1 + 𝑅2 Activity 2. Parallel resistor circuit. 1. Electrical current with voltage │3,0 ± 0,1│V. a. Through R1. 𝑉

3,0 V

I1.1 = 𝑅1 = 100 ohm = 0,03 A 1

So that, 𝑉

I1.1 = 𝑅1

1

I1.1 = V1 × R1-1 𝜕𝐼

𝜕𝐼

dI1.1 = | 𝜕𝑉1.1 d𝑉1 | + | 𝜕𝑅1.1 d𝑅1 | 1

1

dI1.1 = |𝑅1 −1 d𝑉1 | + |𝑉1 × 𝑅1 −2 d𝑅1 | d𝐼1.1 𝐼1.1 d𝐼1.1 𝐼1.1 ∆𝐼1.1 𝐼1.1

𝑅1 −1

=|

𝐼1.1

d𝑉1.1

=|

𝑉1.1

𝑉1 ×𝑅1 −2

d𝑉1 | + |

d𝑅1 |

d𝑅

|+|𝑅1|

∆𝑉

1

= | 𝑉1 +

∆𝑅1

∆𝑉

∆𝑅1

1

∆𝐼1.1= | 𝑉1 + 1

𝐼1.1

𝑅1

𝑅1

| | 𝐼1.1

Because R1 is constant, so that;

∆𝑉

∆𝐼1.1= | 𝑉1 | 𝐼1.1 1

0,1 V

∆𝐼1.1= | 3,0 V | 0,03 A ∆𝐼1.1= 0,033 × 0,03 A ∆𝐼1.1= 0,00099 A RE =

∆𝐼1.1 𝐼1.1

× 100 % =

0,00099 A 0,03 A

× 100 % = 3,3 % (3 Significant Figures)

Reporting of Physics: PR = │𝐼1.1 ± ∆𝐼1.1│ 𝐼1.1 = │0,030000 ± 0,000990│A 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,030000 A – 0,02 A

=| =|

0,025 A 0,01 A

0,025 A

|× 100 %

| × 100 %

| × 100 %

= 40 % b. Through R2. 𝑉

3,0 V

I1.2 = 𝑅1 = 150 ohm = 0,02 A 2

So that, ∆𝑉

∆𝐼1.2= | 𝑉1 | 𝐼1.2 1

0,1 V

∆𝐼1.2= | 3,0 V | 0,02 A ∆𝐼1.2= 0,033 × 0,02 A ∆𝐼1.2= 0,00066 A RE =

∆𝐼1.2 𝐼1.2

× 100 % =

0,00066 A

Reporting of Physics: PR = │𝐼1.2 ± ∆𝐼1.2│

0,03 A

× 100 % = 3,3 % (3 Significant Figures)

𝐼1.2 = │0,020000 ± 0,000660│A 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,020000 A – 0,02 A

=|

0A

| × 100 %

| × 100 %

0A

= |0,02 A| × 100 % =0% The total electric current through is; Itotal.1 = I1.1 + I1.2 Itotal.1 = 0,030000 A + 0,020000 A Itotal.1 = 0,050000 A So that, Itotal.1 = I1.1 + I1.2 𝜕𝐼𝑡𝑜𝑡𝑎𝑙.1

dItotal.1 = |

𝜕𝐼1.2

𝜕𝐼𝑡𝑜𝑡𝑎𝑙.1

d𝐼1.1 | + |

𝜕𝐼1.2

d𝐼1.2 |

dItotal.1 = |1 d𝐼1.1 | + |1 d𝐼1.2 | ∆Itotal.1 = |∆𝐼1.1 | + |∆𝐼1.2 | ∆Itotal.1 = 0,000990 A + 0,000660 A ∆Itotal.1 = 0,001650 A RE =

∆𝐼𝑡𝑜𝑡𝑎𝑙.1 𝐼𝑡𝑜𝑡𝑎𝑙.1

0,001650 A

× 100 % = 0,050000 A × 100 % = 3,3 % (3 Significant Figures)

Reporting of Physics: PR = │𝐼𝑡𝑜𝑡𝑎𝑙.1 ± ∆Itotal.1│ 𝐼𝑡𝑜𝑡𝑎𝑙.1 = │0,05000 ± 0,00165│A 2. Electrical current with voltage │6,0 ± 0,1│V. a. Through R1. 𝑉

6,0 V

I2.1 = 𝑅2 = 100 ohm = 0,06 A 1

So that, ∆𝑉

∆𝐼2.1= | 𝑉2 | 𝐼2.1 2

0,1 V

∆𝐼2.1= | 6,0 V | 0,06 A ∆𝐼2.1= 0,017 × 0,06 A ∆𝐼2.1= 0,00102 A RE =

∆𝐼2.1 𝐼2.1

× 100 % =

0,00102 A 0,06 A

× 100 % = 1,7 % (3 Significant Figures)

Reporting of Physics: PR = │𝐼2.1 ± ∆𝐼2.1│ 𝐼2.1 = │0,06000 ± 0,00102│A 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝐼𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,06000 A – 0,05 A

=|

0,055 A

| × 100 %

| × 100 %

0,01 A

= |0,055 A| × 100 % = 18 % b. Through R2. 𝑉

6,0 V

I2.2 = 𝑅2 = 150 ohm = 0,04 A 2

So that, ∆𝑉

∆𝐼2.2= | 𝑉2 | 𝐼2.2 2

0,1 V

∆𝐼2.2= | 6,0 V | 0,04 A ∆𝐼2.2= 0,017 × 0,02 A ∆𝐼2.2= 0,00068 A RE =

∆𝐼2.2 𝐼2.2

× 100 % =

0,00068 A 0,04 A

× 100 % = 1,7 % (3 Significant Figures)

Reporting of Physics: PR = │𝐼2.2 ± ∆𝐼2.2│ 𝐼1.2 = │0,040000 ± 0,000680│A

𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,040000 A – 0,04 A

=|

0,04 A

| × 100 %

| × 100 %

0A

= |0,04 A| × 100 % =0% The total electric current through is; Itotal.2 = I2.1 + I2.2 Itotal.2 = 0,06000 A + 0,040000 A Itotal.2 = 0,100000 A So that, ∆Itotal.2 = |∆𝐼2.1 | + |∆𝐼2.2 | ∆Itotal.2 = 0,00102 A + 0,000680 A ∆Itotal.2 = 0,0017 A RE =

∆𝐼𝑡𝑜𝑡𝑎𝑙.2 𝐼𝑡𝑜𝑡𝑎𝑙.2

0,0017 A

× 100 % = 0,100000 A × 100 % = 1,7 % (3 Significant Figures)

Reporting of Physics: PR = │𝐼𝑡𝑜𝑡𝑎𝑙.2 ± ∆Itotal.2│ 𝐼𝑡𝑜𝑡𝑎𝑙.2 = │0,10000 ± 0,00170│A 3. Electrical current with voltage │9,0 ± 0,1│V. a. Through R1. 𝑉

9,0 V

I3.1 = 𝑅3 = 100 ohm = 0,09 A 1

So that, ∆𝑉

∆𝐼3.1= | 𝑉3 | 𝐼3.1 3

0,1 V

∆𝐼3.1= | 9,0 V | 0,09 A ∆𝐼3.1= 0,011 × 0,09 A ∆𝐼3.1= 0,001 A RE =

∆𝐼3.1 𝐼3.1

× 100 % =

0,001 A 0,09 A

× 100 % = 1,1 % (3 Significant Figures)

Reporting of Physics: PR = │𝐼3.1 ± ∆𝐼3.1│ 𝐼3.1 = │0,09000 ± 0,00100│A 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,09000 A – 0,08 A

=|

0,085 A

| × 100 %

| × 100 %

0,01 A

= |0,085 A| × 100 % = 11,76 % b. Through R2. 𝑉

9,0 V

I3.2 = 𝑅3 = 150 ohm = 0,06 A 2

So that, ∆𝑉

∆𝐼3.2= | 𝑉3 | 𝐼3.2 3

0,1 V

∆𝐼3.2= | 9,0 V | 0,06 A ∆𝐼3.2= 0,011 × 0,06 A ∆𝐼3.2= 0,00066 A RE =

∆𝐼3.2 𝐼3.2

× 100 % =

0,00066 A 0,06 A

× 100 % = 1,1 % (3 Significant Figures)

Reporting of Physics: PF = │𝐼3.2 ± ∆𝐼3.2│ 𝐼3.2 = │0,060000 ± 0,000660│A 𝐼

%diff = | 𝑡ℎ𝑒𝑜𝑟𝑦𝐼

− 𝐼𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡

𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0,060000 A – 0,06 A

=|

0,06 A 0A

= |0,06 A| × 100 % =0%

| × 100 %

| × 100 %

The total electric current through is; Itotal.3 = I3.1 + I3.2 Itotal.3 = 0,09000 A + 0,060000 A Itotal.3 = 0,150000 A So that, ∆Itotal.3 = |∆𝐼3.1 | + |∆𝐼3.2 | ∆Itotal.3 = 0,00100 A + 0,000660 A ∆Itotal.3 = 0,00166 A RE =

∆𝐼𝑡𝑜𝑡𝑎𝑙.3 𝐼𝑡𝑜𝑡𝑎𝑙.3

0,00166 A

× 100 % = 0,150000 A × 100 % = 1,11 % (3 Significant Figures)

Reporting of Physics: PR = │𝐼𝑡𝑜𝑡𝑎𝑙.3 ± ∆Itotal.3│ 𝐼𝑡𝑜𝑡𝑎𝑙.3 = │0,15000 ± 0,00166│A Thus, it can be knowed that; 𝐼𝑡𝑜𝑡 = 𝐼1 + 𝐼2 𝑉𝑡𝑜𝑡 = 𝑉1 = 𝑉2 𝐼𝑡𝑜𝑡 =

𝑉 𝑉 + 𝑅1 𝑅2

𝑉 1 1 = 𝑉( + ) 𝑅𝑡𝑜𝑡 𝑅1 𝑅2 1 1 1 =( + ) 𝑅𝑡𝑜𝑡 𝑅1 𝑅2 DISCUSSION Based on observations on one activity that experimental tools are arranged in series, the resistor serves as a voltage divider. Wherein if the value of the voltage across the resistor first and second resistors are added then the result is equal to the voltage source. The results obtained on the theoretical value of the voltage different from the voltage value of the practicum are │4,00 ± 0,25│V with electrical current of │0,016 ± 0,001│ A, │8,000 ± 0,250│V with electrical current of │0,032 ± 0,001│ A, and │12,000 ± 0,248│V with electrical current of │0,048 ± 0,001│ A. However, the results obtained in theory is not too different

from the results obtained during the practicum. And the strong results obtained in the current first activity both strong current value before the first resistor, between the first resistor and a second resistor, and after the first resistor result is always the same. From these results it can be determined that the barriers to substitute a series circuit is Rtotal = R1 + R2 + . . . . While the activities of 2, experimental tools are arranged in parallel, the resistor serves as a flow divider. Where the value of a electric current through the first resistor and the second resistor if the results are summed together with electric currents before branching point (electric currents in total). The results obtained on the value of the total electric current strength in this experiment are │0,05000 ± 0,00165│A with a voltage of │3,0 ± 0,1│V, │0,10000 ± 0,00170│A with a voltage of │6,0 ± 0,1│V, and│0,15000 ± 0,00166│A with a voltage of │9,0 ± 0,1│V. From these results it can be determined that the replacement string 1

1

1

1

of parallel barriers is 𝑅 = 𝑅 + 𝑅 + 𝑅 + . . . . 𝑒𝑞

1

2

3

CONCLUSION AND SUGGESTION 1. Conclusion. a. To assemble a resistor into series and parallel arrangement is by paying attention to the connecting cable that connects the resistor, power supply and basicmeter. If the connecting cable branching, then the resistor circuit called a parallel circuit resistor. While the resistor series circuit, the connecting cable is not branched. b. One way to use basicmeter correctly is by connecting the power supply to the attention basicmeter cargo. c. Principle of the first Kirchoff law is a strong electric current through each resistor is measured, then it will have a value equal to the total current before branching point. d. Characteristics resistor series circuit is functioning as a voltage divider resistors, while the characteristics of the parallel circuit is resistor divider serves as an electric current.

2. Suggestion. a. Apprentice should be more careful and cautious at the time of data collection is done in order to get accurate and the results of its analysis in accordance with theory. b. Assistant should not leave the apprentice when practicum takes place. c. The laboratory manager should separate the broken tools with good tool that is most effective in practice. REFERENCE Bakri, A. Haris., dkk. 2008. Dasar-Dasar Elektronika. Makassar: Badan Penerbit Universitas Negeri Makassar. Herman. dan LFD, Asisten. 2015. Penuntun Praktikum Fisika Dasar 2. Makassar: Laboratorium Fisika Dasar FMIPA UNM. Tipler, Paul A. 2001. Fisika untuk Sains dan Teknik Edisi Ketiga Jilid 2(Terjemahan). Jakarta: Erlangga.