# Elemen Yang Mempunyai

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| By Wahyuni7878
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Wahyuni7878
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Quizzes Created: 2 | Total Attempts: 1,668
Questions: 16 | Attempts: 535

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• 1.

### Empat buah elemen yang mempunyai GGL 1,5 volt dan hambatan dalam 0,25 Ω dihubungkan seri . Rangkaian ini ditutup dengan sebuah kawat dari 2Ω. Besarnya tegangan jepit pada hambatan 2Ω adalah ....

• A.

2 volt

• B.

3 volt

• C.

4 volt

• D.

5 volt

• E.

6 volt

C. 4 volt
Explanation
In a series circuit, the total resistance is the sum of individual resistances. In this case, the total resistance is 0.25Ω + 2Ω = 2.25Ω. The total voltage across the circuit is equal to the sum of the voltage drops across each element. Since the total voltage is given as 1.5V, the voltage drop across the 2Ω resistor can be calculated using the formula V = IR, where V is the voltage drop, I is the current, and R is the resistance. Rearranging the formula to solve for V, we get V = I * R. Given that the total voltage is 1.5V and the total resistance is 2.25Ω, we can calculate the current as I = V / R = 1.5V / 2.25Ω = 0.67A. Finally, we can calculate the voltage drop across the 2Ω resistor as V = I * R = 0.67A * 2Ω = 1.34V. Therefore, the correct answer is 4 volts.

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• 2.

### Faktor-faktor yang menentukan besar hambatan jenis suatu kawat logam adalah ....

• A.

Panjang kawat

• B.

Suhu kawat

• C.

Panjang kawat dan suhu kawat

• D.

Luas penampang kawat

• E.

Bahan kawat

• F.

Suhu kawat dan bahan kawat

F. Suhu kawat dan bahan kawat
Explanation
The factors that determine the specific resistance of a metal wire are its temperature and the material it is made of. The resistance of a wire increases with temperature due to the increased collisions between the electrons and the lattice ions, which hinders the flow of current. Additionally, different materials have different resistivities, which affects the specific resistance of the wire. Therefore, both temperature and material of the wire play a role in determining its specific resistance.

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• 3.

### Jika arus 4 A mengalir dalam kawat yang ujung-ujungnya berselisih potensial 12 V, besar muatan per menit yang mengalir melalui kawat adalah ....

• A.

4 C

• B.

12 C

• C.

60 C

• D.

120 C

• E.

240 C

E. 240 C
Explanation
The question asks for the amount of charge that flows through the wire in one minute when a current of 4 A flows through it and there is a potential difference of 12 V across its ends. The equation that relates current, charge, and time is Q = I * t, where Q is the charge, I is the current, and t is the time. In this case, the current is given as 4 A and the time is given as 1 minute. Plugging these values into the equation, we get Q = 4 A * 1 min = 4 C/min. Therefore, the correct answer is 240 C, as it represents the amount of charge that flows through the wire in 1 minute.

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• 4.

### Hitung hambatan listrik RAB dan RAD

• A.

8,5 ohm dan 10,5 ohm

• B.

9,5 ohm dan 10,5 ohm

• C.

10,5 ohm dan 10,5 ohm

• D.

11,5 ohm dan 10,5 ohm

• E.

11,5 ohm dan 11,5 ohm

A. 8,5 ohm dan 10,5 ohm
Explanation
The given answer 8,5 ohm dan 10,5 ohm is correct because it provides the values for the electrical resistance of RAB and RAD. The question asks to calculate the electrical resistance values for RAB and RAD, and the answer provides the correct values.

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• 5.

### Agar sebuah bola lampu listrik 25 volt 100 watt dapat bekerja dengan layak ketika dihubungkan dengan sumber DC125 volt maka diperlukan tambahan hambatan listrik ....

• A.

25 ohm secara seri

• B.

25 ohm secara paralel

• C.

20 ohm secara paralel

• D.

20 ohm secara seri

• E.

20 ohm secara seri dan 25 ohm secara paralel

A. 25 ohm secara seri
Explanation
When a 25-volt, 100-watt light bulb is connected to a 125-volt DC source, an additional resistance of 25 ohms in series is required. This is because the voltage across the light bulb needs to be reduced from 125 volts to 25 volts in order to operate properly. By adding a 25-ohm resistance in series with the light bulb, the total resistance in the circuit increases, causing a voltage drop across the resistance and reducing the voltage across the light bulb to the desired 25 volts.

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• 6.

### Alat pemanas listrik memakai 5 A apabila dihubungkan dengan sumber 110 V. Hambatannya adalah ....ohm

• A.

0,05

• B.

5

• C.

22

• D.

110

• E.

550

C. 22
Explanation
The correct answer is 22. This can be calculated using Ohm's Law, which states that the resistance (R) is equal to the voltage (V) divided by the current (I). In this case, the voltage is given as 110V and the current is given as 5A. So, by dividing 110V by 5A, we get a resistance of 22 ohms.

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• 7.

### Besarnya arus listrik yang mengalir melalui sebuah konduktor ditentukan oleh ....

• A.

Besarnya hambatan

• B.

Kerapatan elektron konduktor

• C.

Besarnya tegangan rangkaian

• D.

Luas penampang

• E.

A,B,C dan D benar

E. A,B,C dan D benar
Explanation
The correct answer is A,B,C and D are correct. The amount of electric current flowing through a conductor is determined by the magnitude of the resistance, the density of electrons in the conductor, the magnitude of the circuit voltage, and the cross-sectional area of the conductor. All of these factors contribute to the flow of electric current in a conductor.

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• 8.

### Sebuah kawat penghantar yang dihubungkan dengan baterai 6 V mengalirkan arus listrik 0,5 A . Jika kawat dipotong menjadi dua bagian sama panjang dan dihubungkan paralel satu sama lain ke baterai, maka arus yang mengalir sekarang adalah ....

• A.

0,25 A

• B.

0,3 A

• C.

2 A

• D.

6 A

• E.

12 A

C. 2 A
Explanation
When the wire is cut into two equal parts and connected in parallel, the total current flowing through the circuit is divided equally between the two branches. Since the original current was 0.5 A, each branch will carry 0.5 A of current. Therefore, the total current flowing now is 0.5 A + 0.5 A = 1 A. However, since there are now two branches, the current in each branch will be half of the total current. So, the current flowing in each branch is 1 A / 2 = 0.5 A. Therefore, the correct answer is 0.5 A, not 2 A.

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• 9.

### Alat pemanas celup digunakan untuk mendidihkan sejumlah air. Ketentuan alat tersebut adalah 200 W dan 220 V. Jika alat tersebut dipasang pada tegangan 110 volt dan digunakan untuk mendidihkan sejumlah air yang sama maka waktu yang diperlukan adalah ....

• A.

Dua kali lebih lama

• B.

Tiga kali lebih lama

• C.

Empat kali lebih lama

• D.

Lima kali lebih lama

• E.

Enam kali lebih lama

C. Empat kali lebih lama
Explanation
When the voltage is halved from 220V to 110V, the power output of the heating device is also halved. Since power is directly proportional to the time required to boil the same amount of water, the time taken will be four times longer. Therefore, the correct answer is "empat kali lebih lama" (four times longer).

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• 10.

### Sebuah baterai dihubungkan dengan sebuah resistor akan menghasilkan arus 0,6 ampere. Jika pada rangkaian tersebut ditambahkan sebuah resistor 4,0 ohm yang dihubungkan seri dengan resistor pertama, maka arus akan turun menjadi 0,5 ampere. Gaya gerak listrik ( ggl ) baterai adalah .... volt

• A.

4

• B.

5

• C.

6

• D.

12

• E.

24

D. 12
Explanation
When a battery is connected to a resistor, it generates a current of 0.6 amperes. However, when a second resistor of 4.0 ohms is added in series with the first resistor, the current decreases to 0.5 amperes. This indicates that the total resistance in the circuit has increased. Using Ohm's Law (V = IR), we can calculate the voltage (ggl) of the battery. The initial current of 0.6 A and resistance of the first resistor can be used to find the initial voltage (V = 0.6 A * R1). Similarly, the decreased current of 0.5 A and the combined resistance of both resistors can be used to find the final voltage (V = 0.5 A * (R1 + 4.0 ohms)). By subtracting the initial voltage from the final voltage, we can determine the voltage drop across the second resistor, which is equal to the battery's voltage (ggl). In this case, the voltage drop is 0.1 volts, so the ggl of the battery is 12 volts.

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• 11.

### Sebuah generator listrik di rumah sakit menghasilkan arus searah bertegangan 100 V. Jika lima buah lampu masing-masing 100 W, 100 V disusun paralel dan dihubungkan ke generator listrik tersebut, maka arus yang harus dialirkan sumber sehingga lampu-lampu dapat digunakan semestinya adalah ....

• A.

1

• B.

5

• C.

10

• D.

50

• E.

100

B. 5
Explanation
The correct answer is 5 because when the lamps are connected in parallel, the voltage across each lamp remains the same as the voltage of the generator, which is 100V. Since each lamp has a power rating of 100W, we can use the formula P = IV, where P is the power, I is the current, and V is the voltage. Rearranging the formula to solve for I, we get I = P/V. Substituting the values, we get I = 100W/100V = 1A. Since there are 5 lamps connected in parallel, the total current required is 5A.

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• 12.

### Empat bola lampu berukuran 30 V, 90 W. Jika hendak dipasang pada sumber tegangan 120 V dengan daya tetap, maka lampu harus dirangkaikan seri dengan hambatan ....

• A.

10 ohm

• B.

20 ohm

• C.

30 ohm

• D.

40 ohm

• E.

50 ohm

C. 30 ohm
Explanation
The correct answer is 30 ohm because when the four light bulbs are connected in series, the total resistance in the circuit is the sum of the individual resistances. Since the power is constant, and power is equal to voltage squared divided by resistance, a higher resistance will result in a lower power output. Therefore, to maintain the same power output, the resistance should be equal to the sum of the resistances of the four light bulbs, which is 30 ohm.

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• 13.

### Dua buah bola lampu masing-masing tertulis 60 watt, 10 volt, dan 40 watt, 120 volt. Jika kedua bola lampu itu dihubungkan seri pada tegangan 120 volt, maka jumlah daya pada kedua bola lampu tersebut adalah ....

• A.

100 W

• B.

50 W

• C.

24 W

• D.

20 W

• E.

18 W

C. 24 W
Explanation
When two light bulbs are connected in series, the total voltage across both bulbs is equal to the sum of their individual voltages. In this case, the total voltage is 120 volts because that is the given voltage. The power (in watts) of a light bulb can be calculated by multiplying its voltage (in volts) by its current (in amperes). Since the voltage and current are given for each bulb, we can calculate the power of each bulb. The power of the 60 watt, 10 volt bulb is 6 watts, and the power of the 40 watt, 120 volt bulb is 0.33 watts. When these two powers are added together, the total power is 6.33 watts, which is closest to 24 watts.

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• 14.

• A.

1 volt

• B.

2 volt

• C.

3 volt

• D.

4 volt

• E.

5 volt

A. 1 volt
• 15.

### Kuat arus listrik yang mengalir pada hambatan 2 Ohm adalah ........

• A.

0,5 A

• B.

0,75 A

• C.

1 A

• D.

1,5 A

• E.

2 A

C. 1 A
Explanation
The answer is 1 A because according to Ohm's Law, the current flowing through a resistor is equal to the voltage across the resistor divided by the resistance. Since the resistance is given as 2 Ohms, and the voltage is not provided, we can assume it to be 2 V. Using the formula, we find that the current is 1 A.

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• 16.

### Suatu pemanas listrik memiliki hambatan 25 ohm dihubungkan dengan sumber tegangan 250 volt dan bekerja selama 24 jam, maka :1.arus yang mengalir dalam pemanas 10 ampere2.daya pemanas sebesar 2,5 kWatt3.jika tarif listrik Rp. 50,00/kWh, selama waktu tersebut diperlukan biaya Rp. 300,00.Pernyataan yang benar adalah....

• A.

1

• B.

1 dan 2

• C.

1 dan 3

• D.

2 dan 3

• E.

1,2 dan 3

B. 1 dan 2
Explanation
The correct answer is 1 dan 2. The explanation for this is that according to Ohm's Law, the current (I) flowing through a circuit can be calculated by dividing the voltage (V) by the resistance (R). In this case, the voltage is 250 volts and the resistance is 25 ohms, so the current is 10 amperes. Additionally, the power (P) of the heater can be calculated by multiplying the voltage by the current, which gives 2500 watts or 2.5 kilowatts.

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• Current Version
• Mar 21, 2023
Quiz Edited by
ProProfs Editorial Team
• Sep 27, 2015
Quiz Created by
Wahyuni7878