Q. What is the formula for the total capacitance of two capacitors in parallel?
A.
C_total = C1 + C2
B.
C_total = C1 * C2
C.
C_total = C1 / C2
D.
C_total = C1 - C2
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Solution
For capacitors in parallel, the total capacitance is the sum: C_total = C1 + C2.
Correct Answer:
A
— C_total = C1 + C2
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Q. What is the frequency of an AC signal with a period of 0.02 seconds?
A.
50 Hz
B.
100 Hz
C.
25 Hz
D.
200 Hz
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Solution
Frequency (f) is the reciprocal of the period (T). Thus, f = 1/T = 1/0.02 s = 50 Hz.
Correct Answer:
B
— 100 Hz
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Q. What is the function of a capacitor in an electric circuit? (2022)
A.
Store energy
B.
Increase current
C.
Decrease voltage
D.
Convert AC to DC
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Solution
A capacitor is used to store electrical energy in an electric field, which can be released when needed.
Correct Answer:
A
— Store energy
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Q. What is the magnetic field around a straight conductor carrying current?
A.
Uniform
B.
Concentric circles
C.
Radial lines
D.
None of the above
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Solution
The magnetic field around a straight conductor carrying current forms concentric circles around the wire.
Correct Answer:
B
— Concentric circles
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Q. What is the magnetic field around a straight current-carrying conductor? (2023)
A.
Uniform
B.
Concentric circles
C.
Radial lines
D.
None of the above
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Solution
The magnetic field around a straight current-carrying conductor forms concentric circles around the wire, as described by Ampère's circuital law.
Correct Answer:
B
— Concentric circles
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Q. What is the magnetic field strength at the center of a circular loop of radius 0.1 m carrying a current of 5 A?
A.
0.1 T
B.
0.2 T
C.
0.5 T
D.
1 T
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Solution
The magnetic field at the center of a circular loop is given by B = (μ₀ * I) / (2 * R). Using μ₀ = 4π x 10^-7 Tm/A, B = (4π x 10^-7 * 5) / (2 * 0.1) = 0.1 T.
Correct Answer:
B
— 0.2 T
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Q. What is the magnetic field strength at the center of a circular loop of radius 0.1 m carrying a current of 3 A?
A.
0.03 T
B.
0.1 T
C.
0.15 T
D.
0.2 T
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Solution
The magnetic field (B) at the center of a circular loop is given by B = (μ₀ * I) / (2 * R). Using μ₀ = 4π x 10^-7 Tm/A, B = (4π x 10^-7 * 3) / (2 * 0.1) = 0.1 T.
Correct Answer:
B
— 0.1 T
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Q. What is the magnetic field strength at the center of a circular loop of radius r carrying a current I?
A.
μ0I/2r
B.
μ0I/r
C.
μ0I/4r
D.
μ0I/πr
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Solution
The magnetic field at the center of a circular loop is given by B = (μ0I)/(2r).
Correct Answer:
B
— μ0I/r
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Q. What is the potential difference across a 10-ohm resistor carrying a current of 3 A?
A.
30 V
B.
10 V
C.
3 V
D.
0.3 V
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Solution
Using Ohm's Law (V = I * R), V = 3 A * 10 Ω = 30 V.
Correct Answer:
A
— 30 V
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Q. What is the power consumed by a device that operates at 12 volts and draws a current of 2 amperes? (2020)
A.
24 W
B.
12 W
C.
6 W
D.
48 W
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Solution
Power (P) is given by P = V * I. Thus, P = 12 V * 2 A = 24 W.
Correct Answer:
A
— 24 W
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Q. What is the power consumed by a device that operates at 120 V and draws a current of 2 A?
A.
240 W
B.
60 W
C.
120 W
D.
30 W
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Solution
Power (P) is calculated using the formula P = V * I. Therefore, P = 120 V * 2 A = 240 W.
Correct Answer:
A
— 240 W
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Q. What is the power consumed by a device that operates at 120 volts and draws a current of 3 amperes?
A.
360 W
B.
40 W
C.
120 W
D.
240 W
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Solution
Power P is calculated using P = V * I. Therefore, P = 120 V * 3 A = 360 W.
Correct Answer:
A
— 360 W
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Q. What is the power consumed by a device that operates at 120V and draws a current of 3A? (2021)
A.
360 W
B.
240 W
C.
120 W
D.
30 W
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Solution
Power (P) = Voltage (V) × Current (I) = 120V × 3A = 360W.
Correct Answer:
A
— 360 W
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Q. What is the power consumed by a device that operates at a voltage of 120 volts and draws a current of 3 amperes?
A.
360 watts
B.
40 watts
C.
120 watts
D.
240 watts
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Solution
Power (P) is calculated as P = V * I. Thus, P = 120 V * 3 A = 360 W.
Correct Answer:
A
— 360 watts
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Q. What is the resistance of a wire of length 2 m and cross-sectional area 1 mm² if the resistivity of the material is 1.7 x 10^-8 Ωm?
A.
0.034 Ω
B.
0.017 Ω
C.
0.068 Ω
D.
0.085 Ω
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Solution
Resistance R = ρ * (L/A). Here, R = (1.7 x 10^-8 Ωm) * (2 m / (1 x 10^-6 m²)) = 0.034 Ω.
Correct Answer:
A
— 0.034 Ω
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Q. What is the total capacitance of two capacitors of 2 µF and 3 µF connected in series? (2023)
A.
1.2 µF
B.
5 µF
C.
6 µF
D.
0.86 µF
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Solution
For capacitors in series, 1/C_total = 1/C1 + 1/C2. Thus, 1/C_total = 1/2 + 1/3 = 5/6, so C_total = 1.2 µF.
Correct Answer:
A
— 1.2 µF
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Q. What is the total capacitance of two capacitors of 3 µF and 6 µF connected in parallel?
A.
2 µF
B.
3 µF
C.
9 µF
D.
18 µF
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Solution
In parallel, the total capacitance is the sum: C_total = C1 + C2 = 3 µF + 6 µF = 9 µF.
Correct Answer:
C
— 9 µF
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Q. What is the total capacitance of two capacitors of 3 µF and 6 µF in parallel?
A.
9 µF
B.
2 µF
C.
18 µF
D.
1.5 µF
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Solution
In parallel, the total capacitance is the sum: C_total = C1 + C2 = 3 µF + 6 µF = 9 µF.
Correct Answer:
A
— 9 µF
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Q. What is the total capacitance of two capacitors of 4 microfarads and 6 microfarads connected in series? (2019)
A.
2.4 microfarads
B.
10 microfarads
C.
24 microfarads
D.
1.5 microfarads
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Solution
For capacitors in series, 1/C_eq = 1/C1 + 1/C2. Thus, 1/C_eq = 1/4 + 1/6 = 5/12. Therefore, C_eq = 2.4 microfarads.
Correct Answer:
A
— 2.4 microfarads
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Q. What is the total capacitance of two capacitors, C1 = 3μF and C2 = 6μF, connected in series? (2022)
A.
2μF
B.
9μF
C.
1.5μF
D.
18μF
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Solution
The formula for total capacitance in series is 1/C_eq = 1/C1 + 1/C2. Thus, 1/C_eq = 1/3 + 1/6 = 1/2, so C_eq = 2μF.
Correct Answer:
A
— 2μF
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Q. What is the total current in a parallel circuit with three branches carrying 2 A, 3 A, and 5 A respectively?
A.
10 A
B.
5 A
C.
3 A
D.
2 A
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Solution
In a parallel circuit, the total current is the sum of the currents in each branch: I_total = 2 A + 3 A + 5 A = 10 A.
Correct Answer:
A
— 10 A
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Q. What is the total current in a parallel circuit with two branches carrying 3 A and 5 A?
A.
8 A
B.
15 A
C.
5 A
D.
3 A
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Solution
In a parallel circuit, the total current is the sum of the currents in each branch: I_total = I1 + I2 = 3 A + 5 A = 8 A.
Correct Answer:
A
— 8 A
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Q. What is the total energy stored in a capacitor of capacitance 20 microfarads charged to 10 volts?
A.
1 joule
B.
0.1 joule
C.
0.01 joule
D.
0.5 joule
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Solution
Energy (E) stored in a capacitor is given by E = 0.5 * C * V^2. Thus, E = 0.5 * 20 x 10^-6 F * (10 V)^2 = 0.1 joules.
Correct Answer:
B
— 0.1 joule
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Q. What is the total resistance in a series circuit with resistors of 4 ohms, 6 ohms, and 10 ohms? (2023)
A.
20 ohms
B.
10 ohms
C.
6 ohms
D.
4 ohms
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Solution
In a series circuit, total resistance (R_total) is the sum of individual resistances: R_total = 4 + 6 + 10 = 20 ohms.
Correct Answer:
A
— 20 ohms
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