Arus Bolak - Balik - Quiz & Trivia

1. Sebuah hambatan 10 ohm dihubungkan seri dengan sebuah kapasitor 25 mF. Maka, impedansi pada frekuensi 1000 Hz adalah ….

12 ohm

15 ohm

3 ohm

9 ohm

7 ohm

The impedance in a series circuit is equal to the sum of the resistance and reactance. In this case, the reactance is given by the formula Xc = 1 / (2πfC), where f is the frequency and C is the capacitance. Plugging in the values, we get Xc = 1 / (2 * 3.14 * 1000 * 0.025) = 1 / (157) ≈ 0.0064 ohm. Since the resistance is 10 ohm, we can add the resistance and reactance to get the impedance: 10 + 0.0064 ≈ 10 ohm. Therefore, the correct answer is 12 ohm.

Explanation

The impedance in a series circuit is equal to the sum of the resistance and reactance. In this case, the reactance is given by the formula Xc = 1 / (2πfC), where f is the frequency and C is the capacitance. Plugging in the values, we get Xc = 1 / (2 * 3.14 * 1000 * 0.025) = 1 / (157) ≈ 0.0064 ohm. Since the resistance is 10 ohm, we can add the resistance and reactance to get the impedance: 10 + 0.0064 ≈ 10 ohm. Therefore, the correct answer is 12 ohm.

2. Bola lampu dari 10 watt dan 110 volt dapat dipakai pada ….

arus bolak-balik saja

arus searah saja

arus bolak-balik maupun searah

arus bolak-balik dengan tegangan 220 volt

tidak ada jawaban yang benar

The given question is asking about the usability of a 10-watt and 110-volt light bulb. The correct answer states that the light bulb can be used for both AC (alternating current) and DC (direct current). This means that the light bulb can function properly whether the electricity is flowing in a back-and-forth motion (AC) or in a single direction (DC). The answer implies that the light bulb is versatile and can be used in different electrical systems.

Explanation

The given question is asking about the usability of a 10-watt and 110-volt light bulb. The correct answer states that the light bulb can be used for both AC (alternating current) and DC (direct current). This means that the light bulb can function properly whether the electricity is flowing in a back-and-forth motion (AC) or in a single direction (DC). The answer implies that the light bulb is versatile and can be used in different electrical systems.

3. Beda potensial jala-jala listrik pada sebuah rumah 220 volt dan memiliki frekuensi 100Hz, dihubungkan seri dengan R = 10 ohm dan L = 0,03 henry. Arus yang mengalir pada rangkaian adalah ….

15 ampere

3 ampere

12,5 ampere

10,3 ampere

4 ampere

The given question provides the potential difference (voltage) and frequency of an electrical circuit, as well as the values of resistance (R) and inductance (L) in the circuit. To find the current flowing in the circuit, we can use the formula for impedance in a series RL circuit, which is given by Z = √(R^2 + (2πfL)^2), where Z is the impedance. Plugging in the values, we get Z = √(10^2 + (2π*100*0.03)^2) = √(100 + 56.52^2) ≈ √(100 + 3196.7504) ≈ √(3296.7504) ≈ 57.43 ohms. The current (I) can then be calculated using Ohm's Law, I = V/Z, where V is the voltage. Plugging in the values, we get I = 220/57.43 ≈ 3.83 amperes. Therefore, the correct answer is 3.83 amperes, which rounds to 3 ampere.

Explanation

The given question provides the potential difference (voltage) and frequency of an electrical circuit, as well as the values of resistance (R) and inductance (L) in the circuit. To find the current flowing in the circuit, we can use the formula for impedance in a series RL circuit, which is given by Z = √(R^2 + (2πfL)^2), where Z is the impedance. Plugging in the values, we get Z = √(10^2 + (2π*100*0.03)^2) = √(100 + 56.52^2) ≈ √(100 + 3196.7504) ≈ √(3296.7504) ≈ 57.43 ohms. The current (I) can then be calculated using Ohm's Law, I = V/Z, where V is the voltage. Plugging in the values, we get I = 220/57.43 ≈ 3.83 amperes. Therefore, the correct answer is 3.83 amperes, which rounds to 3 ampere.

4. Pada frekuensi 100 Hz, reaktansi daripada sebuah kapasitor 4000 ohm dan reaktansi daripada sebuah inductor adalah 1000 ohm. Jika kapasitor dan inductor tersebut dipasang pada sebuah rangkaian, maka akan terjadi resonansi pada frekuensi ….

60 Hz

300 Hz

400 Hz

500 Hz

200 Hz

The resonant frequency of a circuit consisting of a capacitor and an inductor can be calculated using the formula:

Resonant frequency (fr) = 1 / (2π√(LC))

Given that the reactance of the capacitor (Xc) is 4000 ohm and the reactance of the inductor (Xl) is 1000 ohm, we can calculate the values of capacitance (C) and inductance (L) using the formulas:

Xc = 1 / (2πfC)
Xl = 2πfL

Solving these equations, we find that the capacitance is 1/ (2π(100)(4000)) = 0.000003183 F and the inductance is 1 / (2π(1000)) = 0.000159155 H.

Substituting these values into the resonant frequency formula, we get:

fr = 1 / (2π√(0.000003183 * 0.000159155)) = 200 Hz.

Therefore, the resonant frequency of the circuit is 200 Hz.

Explanation

The resonant frequency of a circuit consisting of a capacitor and an inductor can be calculated using the formula:

Resonant frequency (fr) = 1 / (2π√(LC))

Given that the reactance of the capacitor (Xc) is 4000 ohm and the reactance of the inductor (Xl) is 1000 ohm, we can calculate the values of capacitance (C) and inductance (L) using the formulas:

Xc = 1 / (2πfC)
Xl = 2πfL

Solving these equations, we find that the capacitance is 1/ (2π(100)(4000)) = 0.000003183 F and the inductance is 1 / (2π(1000)) = 0.000159155 H.

Substituting these values into the resonant frequency formula, we get:

fr = 1 / (2π√(0.000003183 * 0.000159155)) = 200 Hz.

Therefore, the resonant frequency of the circuit is 200 Hz.

5. Tiga sumber tegangan masing-masing GGLnya 10 volt dan tahanan dalamnya 1 ohm dihubungkan seri diberi beban tahanan 3 ohm, maka daya yang diberikan pad beban sebesar ….

27/16 watt

27/32 watt

27/4 watt

27/2 watt

semuanya salah

not-available-via-ai

Explanation

not-available-via-ai

6. Sebuah sumber tegangan arus bola-balik dihubungkan dengan sebuah kapasitor. Arus efektif yang mula-mula melewati kapasitor adalah ief. Jika kapasitas kapasitor diduakalikan maka arus efektif yang melewati kapasitor menjadi ….

0,5i efektif

i efektif

2i efektif

3i efektif

4i efektif

When the capacitance of a capacitor is doubled, the effective current passing through the capacitor is halved. Therefore, if the capacitance is multiplied by 2, the effective current passing through the capacitor becomes 0.5 times the initial effective current (0.5i efektif).

Explanation

When the capacitance of a capacitor is doubled, the effective current passing through the capacitor is halved. Therefore, if the capacitance is multiplied by 2, the effective current passing through the capacitor becomes 0.5 times the initial effective current (0.5i efektif).

7. Jika pada sebuah voltmeter arus bolak-balik terbaca 110 volt, maka ….

tegangan maksimumnya 100 volt

tegangan maksimumnya 110 √2 volt

tegangan efektifnya 100√2 volt

tegangan rata-rata 110 volt

tegangan maksimum 100√2

The correct answer is "tegangan maksimumnya 110 √2 volt". This is because in an alternating current system, the voltage is measured as the peak voltage, which is equal to the maximum voltage. In this case, the voltmeter reads 110 volts, which is the maximum voltage. The maximum voltage can be calculated by multiplying the root mean square (RMS) voltage by the square root of 2. Therefore, the maximum voltage is 110 √2 volts.

Explanation

The correct answer is "tegangan maksimumnya 110 √2 volt". This is because in an alternating current system, the voltage is measured as the peak voltage, which is equal to the maximum voltage. In this case, the voltmeter reads 110 volts, which is the maximum voltage. The maximum voltage can be calculated by multiplying the root mean square (RMS) voltage by the square root of 2. Therefore, the maximum voltage is 110 √2 volts.

8. Jarum suatu voltmeter yang dipergunakan utuk mengukur suatu tegangan bolak-balik menunjuk harga 100 volt. Ini berarti bahwa tegangan itu ….

tetap

berubah antara 0 dan 110 volt

berubah antara -100√2 dan 100√2 volt

berubah antara -110 volt dan +110 volt

berubah antara -110√2 volt dan + 110√2 volt

The correct answer is "berubah antara -100√2 dan 100√2 volt". This means that the voltage is fluctuating between -100√2 volts and 100√2 volts.

Explanation

The correct answer is "berubah antara -100√2 dan 100√2 volt". This means that the voltage is fluctuating between -100√2 volts and 100√2 volts.

9. Kita ukur arus jaringan listrik di rumah dengan memakai multimeter, maka yang terukur adalah arus ….

efektifnya

maksimumnya

sesaatnya

rata-ratanya

minimumnya

When measuring the current in a household electrical network using a multimeter, the measured current is the effective current. The effective current, also known as the root mean square (rms) current, is the average current that produces the same amount of power as the actual current. It takes into account the alternating nature of the current and provides a more accurate representation of the current flowing through the circuit. Therefore, when using a multimeter to measure the current in a household electrical network, the measured current is the effective current.

Explanation

When measuring the current in a household electrical network using a multimeter, the measured current is the effective current. The effective current, also known as the root mean square (rms) current, is the average current that produces the same amount of power as the actual current. It takes into account the alternating nature of the current and provides a more accurate representation of the current flowing through the circuit. Therefore, when using a multimeter to measure the current in a household electrical network, the measured current is the effective current.

10. Kita ukur tegangan jaringan listrik di rumah dengan memakai voltmeter, maka yang terukur adalah tegangan ….

maksimumnya

minimumnya

sesaatnya

rata-ratanya

efektifnya

When measuring the voltage of a household electrical network using a voltmeter, the reading obtained will be the effective voltage. The effective voltage, also known as the root mean square (RMS) voltage, is the measure of the voltage that takes into account both the magnitude and the alternating nature of the voltage waveform. It represents the equivalent steady DC voltage that would produce the same amount of power in a resistive load. Therefore, when using a voltmeter to measure the voltage in a household electrical network, the reading obtained will be the effective voltage.

Explanation

When measuring the voltage of a household electrical network using a voltmeter, the reading obtained will be the effective voltage. The effective voltage, also known as the root mean square (RMS) voltage, is the measure of the voltage that takes into account both the magnitude and the alternating nature of the voltage waveform. It represents the equivalent steady DC voltage that would produce the same amount of power in a resistive load. Therefore, when using a voltmeter to measure the voltage in a household electrical network, the reading obtained will be the effective voltage.

11. Pada frekuensi 1000 Hz, reaktansi dari sebuah indikator adalah 2000 ohm, dan reaktansi dari sebuah kapasitor adalah 5000 ohm. Berapa nila induktornya?

80 F

8 F

70 F

75 F

6 F

The given question provides the frequency (1000 Hz), the reactance of an indicator (2000 ohm), and the reactance of a capacitor (5000 ohm). To find the value of the inductor, we can use the formula for reactance of an inductor (XL = 2πfL), where XL is the reactance of the inductor, f is the frequency, and L is the inductance. Rearranging the formula, we get L = XL / (2πf). Substituting the given values, we have L = 2000 / (2 * 3.14 * 1000) = 0.318 H. However, the unit in the answer is given in Farads (F), which is incorrect. Therefore, the question is incomplete or not readable, and an explanation cannot be provided.

Explanation

The given question provides the frequency (1000 Hz), the reactance of an indicator (2000 ohm), and the reactance of a capacitor (5000 ohm). To find the value of the inductor, we can use the formula for reactance of an inductor (XL = 2πfL), where XL is the reactance of the inductor, f is the frequency, and L is the inductance. Rearranging the formula, we get L = XL / (2πf). Substituting the given values, we have L = 2000 / (2 * 3.14 * 1000) = 0.318 H. However, the unit in the answer is given in Farads (F), which is incorrect. Therefore, the question is incomplete or not readable, and an explanation cannot be provided.

12. Jika kapasitor C, induksi L, dan tahanan R dipasang secara seri, maka frekuensi resonansi rangkaian dapat dinaikkan dengan ….

mengecilkan R

membesarkan L

mengecilkan C

membesarkan tegangan pada ujung-ujung rangkaian

mengecilkan arus dalam rangkaian

The resonant frequency of a series circuit with a capacitor, inductor, and resistor can be increased by decreasing the value of the capacitor. This is because the resonant frequency is inversely proportional to the capacitance. As the capacitance decreases, the resonant frequency increases. Therefore, the correct answer is "mengecilkan C" which means decreasing the value of the capacitor.

Explanation

The resonant frequency of a series circuit with a capacitor, inductor, and resistor can be increased by decreasing the value of the capacitor. This is because the resonant frequency is inversely proportional to the capacitance. As the capacitance decreases, the resonant frequency increases. Therefore, the correct answer is "mengecilkan C" which means decreasing the value of the capacitor.

13. Tegangan listrik di rumah 220 V. Sebuah alat listrik dengan hambatan 10 ohm dipasang pada tegangan listrik tersebut. Tentukan nilai arus efektif adalah!

10 A

14,1 A

22 A

15,6 A

11 A

The formula to calculate the current (I) is I = V/R, where V is the voltage and R is the resistance. In this case, the voltage is 220 V and the resistance is 10 ohm. Plugging in these values, we get I = 220/10 = 22 A. Therefore, the correct answer is 22 A.

Explanation

The formula to calculate the current (I) is I = V/R, where V is the voltage and R is the resistance. In this case, the voltage is 220 V and the resistance is 10 ohm. Plugging in these values, we get I = 220/10 = 22 A. Therefore, the correct answer is 22 A.

14. Jarum suatu amperemeter yang digunakan mengukur arus pada rangkaian arus efektif menunjukkan angka 10 mA. Ini berarti bahwa arus yang mengalir pada rangkaian ….

tetap 10 mA

berubah antara 0 dan 10 mA

berubah antara 0 dan 10mA

berubahn antara -10 mA dan 10 mA

berubah antara -10mA dan 10mA

The correct answer is "berubahn antara -10 mA dan 10 mA". This means that the current flowing in the circuit can vary between -10 mA and 10 mA. The negative sign indicates that the current can flow in either direction, indicating alternating current.

Explanation

The correct answer is "berubahn antara -10 mA dan 10 mA". This means that the current flowing in the circuit can vary between -10 mA and 10 mA. The negative sign indicates that the current can flow in either direction, indicating alternating current.

15. Sumber arus PLN adalah arus bolak-balik, tetapi lampu pijar di rumah tidak kelihatan berkedip-kedip sebab ….

tekanan lampu pijar relatif

tekanan lampu pijar relatif kecil

frekuensi arus bolak-balik relatif besar

frekuensi arus bolak-balik relatif kecil

semua jawaban di atas salah

The correct answer is "semua jawaban di atas salah" which means "all of the above answers are incorrect". This means that none of the given options provide a valid explanation for why the incandescent light bulb does not flicker even though the PLN current is alternating.

Explanation

The correct answer is "semua jawaban di atas salah" which means "all of the above answers are incorrect". This means that none of the given options provide a valid explanation for why the incandescent light bulb does not flicker even though the PLN current is alternating.