Q. In a Carnot engine, what does the efficiency depend on?
A.
The temperature of the hot reservoir
B.
The temperature of the cold reservoir
C.
Both temperatures
D.
None of the above
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Solution
The efficiency of a Carnot engine depends on both the temperature of the hot reservoir and the cold reservoir.
Correct Answer: C — Both temperatures
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Q. In a Carnot engine, what is the efficiency dependent on?
A.
The work done
B.
The temperatures of the hot and cold reservoirs
C.
The type of working substance
D.
The volume of the gas
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Solution
The efficiency of a Carnot engine is dependent on the temperatures of the hot and cold reservoirs, given by η = 1 - (Tc/Th).
Correct Answer: B — The temperatures of the hot and cold reservoirs
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Q. In a Carnot engine, which of the following is true?
A.
It operates between two temperatures
B.
It is 100% efficient
C.
It can operate with any working substance
D.
It is a perpetual motion machine
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Solution
A Carnot engine operates between two temperatures, absorbing heat from a hot reservoir and rejecting heat to a cold reservoir.
Correct Answer: A — It operates between two temperatures
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Q. In a chemical reaction, if the enthalpy change is positive, the reaction is classified as:
A.
Exothermic
B.
Endothermic
C.
Isothermal
D.
Adiabatic
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Solution
A positive enthalpy change indicates that the reaction absorbs heat, classifying it as endothermic.
Correct Answer: B — Endothermic
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Q. In a chemical reaction, if the enthalpy of products is less than that of reactants, what can be concluded?
A.
The reaction is endothermic
B.
The reaction is exothermic
C.
The reaction is at equilibrium
D.
The reaction is spontaneous
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Solution
If the enthalpy of products is less, the reaction releases heat, indicating it is exothermic.
Correct Answer: B — The reaction is exothermic
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Q. In a circuit with a 10V battery and two resistors (3Ω and 6Ω) in series, what is the voltage drop across the 6Ω resistor?
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Solution
Voltage drop across R2 = (R2 / (R1 + R2)) * Vtotal = (6 / (3 + 6)) * 10 = 6.67V.
Correct Answer: C — 6V
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Q. In a circuit with a 10V battery and two resistors in parallel (4Ω and 6Ω), what is the voltage across each resistor?
A.
10V for both
B.
5V for both
C.
10V for 4Ω and 6V for 6Ω
D.
10V for 6Ω and 4V for 4Ω
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Solution
In a parallel circuit, the voltage across each resistor is the same as the source voltage. Therefore, both resistors have 10V across them.
Correct Answer: A — 10V for both
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Q. In a circuit with a 10V battery and two resistors in series (2Ω and 3Ω), what is the voltage drop across the 3Ω resistor?
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Solution
Using the voltage divider rule, V3 = (R3 / (R2 + R3)) * Vtotal = (3 / (2 + 3)) * 10 = 6V.
Correct Answer: B — 6V
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Q. In a circuit with a 10V battery and two resistors in series (R1 = 2Ω, R2 = 3Ω), what is the current flowing through the circuit?
A.
2A
B.
1A
C.
0.5A
D.
3A
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Solution
Total resistance R_total = R1 + R2 = 2Ω + 3Ω = 5Ω. Current I = V / R_total = 10V / 5Ω = 2A.
Correct Answer: B — 1A
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Q. In a circuit with a 10V battery and two resistors in series (R1 = 2Ω, R2 = 8Ω), what is the current flowing through the circuit?
A.
0.5A
B.
1A
C.
2A
D.
5A
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Solution
Total resistance R_total = R1 + R2 = 2Ω + 8Ω = 10Ω. Current I = V/R = 10V/10Ω = 1A.
Correct Answer: B — 1A
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Q. In a circuit with a 10V battery and two resistors of 5 ohms each in series, what is the current flowing through the circuit?
A.
1 A
B.
2 A
C.
0.5 A
D.
5 A
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Solution
Total resistance R_total = 5 + 5 = 10 ohms. Current I = V/R_total = 10V / 10 ohms = 1 A.
Correct Answer: B — 2 A
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Q. In a circuit with a 10V battery and two resistors of 5Ω each in series, what is the voltage across the second resistor?
A.
5V
B.
10V
C.
2.5V
D.
0V
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Solution
In series, the voltage is divided equally: V2 = (R2 / (R1 + R2)) * Vtotal = (5 / (5 + 5)) * 10 = 5V.
Correct Answer: A — 5V
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Q. In a circuit with a 12 V battery and two resistors in series (4 ohms and 8 ohms), what is the voltage drop across the 8 ohm resistor?
A.
4 V
B.
8 V
C.
6 V
D.
12 V
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Solution
The total resistance is 4 + 8 = 12 ohms. The current I = V / R = 12 V / 12 ohms = 1 A. The voltage drop across the 8 ohm resistor is V = I * R = 1 A * 8 ohms = 8 V.
Correct Answer: B — 8 V
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Q. In a circuit with a 12V battery and two resistors in parallel (4 ohms and 6 ohms), what is the total current supplied by the battery?
A.
3 A
B.
2 A
C.
1 A
D.
4 A
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Solution
First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 = 5/12, so R_eq = 12/5 = 2.4 ohms. Then, using Ohm's law, I = V/R, we have I = 12V / 2.4 ohms = 5 A.
Correct Answer: A — 3 A
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Q. In a circuit with a 12V battery and two resistors in series (3 ohms and 6 ohms), what is the voltage drop across the 6 ohm resistor?
A.
8 V
B.
4 V
C.
6 V
D.
12 V
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Solution
The total resistance is 9 ohms. The current is I = V/R = 12V / 9 ohms = 4/3 A. The voltage drop across the 6 ohm resistor is V = I * R = (4/3 A) * 6 ohms = 8 V.
Correct Answer: A — 8 V
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Q. In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is the current flowing through the circuit?
A.
0.5A
B.
1A
C.
1.5A
D.
2A
Show solution
Solution
Total resistance R = R1 + R2 = 4Ω + 8Ω = 12Ω. Using Ohm's law, I = V/R = 12V / 12Ω = 1A.
Correct Answer: B — 1A
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Q. In a circuit with a 12V battery and two resistors in series (R1 = 4Ω, R2 = 8Ω), what is the voltage across R2?
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Solution
The total resistance is R_total = R1 + R2 = 4Ω + 8Ω = 12Ω. The current is I = V/R_total = 12V / 12Ω = 1A. The voltage across R2 is V_R2 = I * R2 = 1A * 8Ω = 8V.
Correct Answer: C — 8V
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Q. In a circuit with a 12V battery and two resistors in series (R1 = 4Ω, R2 = 8Ω), what is the current flowing through the circuit?
A.
0.5A
B.
1A
C.
1.5A
D.
2A
Show solution
Solution
Total resistance R_total = R1 + R2 = 4Ω + 8Ω = 12Ω. Current I = V/R_total = 12V/12Ω = 1A.
Correct Answer: B — 1A
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Q. In a circuit with a 12V battery and two resistors of 4 ohms and 8 ohms in series, what is the voltage drop across the 8 ohm resistor?
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Solution
Total resistance R_total = 4 + 8 = 12 ohms. Current I = V/R_total = 12V/12 ohms = 1A. Voltage drop across 8 ohm resistor = I * R = 1A * 8 ohms = 8V.
Correct Answer: B — 6V
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Q. In a circuit with a 12V battery and two resistors of 4Ω and 8Ω in series, what is the voltage across the 8Ω resistor?
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Solution
Using the voltage divider rule, V = (R2 / (R1 + R2)) * Vtotal = (8 / (4 + 8)) * 12 = 8V.
Correct Answer: C — 8V
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Q. In a circuit with a 12V battery and two resistors of 4Ω and 8Ω in series, what is the current flowing through the circuit?
A.
0.5A
B.
1A
C.
1.5A
D.
2A
Show solution
Solution
Total resistance R = R1 + R2 = 4Ω + 8Ω = 12Ω. Current I = V/R = 12V / 12Ω = 1A.
Correct Answer: B — 1A
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Q. In a circuit with a 15V battery and three resistors (5Ω, 10Ω, and 15Ω) in parallel, what is the equivalent resistance?
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Solution
1/R_eq = 1/5 + 1/10 + 1/15 = 0.1 + 0.0667 + 0.0667 = 0.2334. Therefore, R_eq = 4.29Ω.
Correct Answer: B — 3Ω
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Q. In a circuit with a 15V battery and two resistors in series (5Ω and 10Ω), what is the voltage drop across the 10Ω resistor?
A.
5V
B.
10V
C.
15V
D.
20V
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Solution
Using the voltage divider rule, V2 = (R2 / (R1 + R2)) * Vtotal = (10 / (5 + 10)) * 15 = 10V.
Correct Answer: B — 10V
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Q. In a circuit with a 15V battery and two resistors in series (R1 = 5Ω, R2 = 10Ω), what is the voltage across R2?
A.
10V
B.
5V
C.
15V
D.
0V
Show solution
Solution
Voltage across R2 = (R2 / (R1 + R2)) * V = (10Ω / (5Ω + 10Ω)) * 15V = 10V.
Correct Answer: A — 10V
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Q. In a circuit with a 20V battery and two resistors (4Ω and 8Ω) in parallel, what is the total current supplied by the battery?
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Solution
Total resistance in parallel, 1/Req = 1/R1 + 1/R2 = 1/4 + 1/8 = 3/8. Therefore, Req = 8/3Ω. Total current I = V/Req = 20V / (8/3) = 7.5A.
Correct Answer: C — 4A
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Q. In a circuit with a 24 V battery and a 6 ohm resistor, what is the current flowing through the resistor?
A.
2 A
B.
4 A
C.
6 A
D.
8 A
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Solution
Using Ohm's law, I = V/R = 24 V / 6 ohms = 4 A.
Correct Answer: B — 4 A
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Q. In a circuit with a 24 V battery and a total resistance of 6 ohms, what is the total current flowing through the circuit?
A.
2 A
B.
4 A
C.
6 A
D.
8 A
Show solution
Solution
Using Ohm's law, I = V/R = 24 V / 6 ohms = 4 A.
Correct Answer: B — 4 A
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Q. In a circuit with a 24V battery and a 6 ohm resistor, what is the voltage drop across the resistor?
A.
4 V
B.
12 V
C.
24 V
D.
6 V
Show solution
Solution
Using Ohm's law, V = I * R. First, find the current I = V/R = 24V / 6 ohms = 4 A. Then, V_drop = I * R = 4 A * 6 ohms = 24 V.
Correct Answer: B — 12 V
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Q. In a circuit with a 24V battery and a total resistance of 8 ohms, what is the total current flowing in the circuit?
A.
3 A
B.
2 A
C.
4 A
D.
6 A
Show solution
Solution
Using Ohm's law, I = V/R = 24V / 8Ω = 3 A.
Correct Answer: C — 4 A
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Q. In a circuit with a 24V battery and three resistors (R1 = 4Ω, R2 = 4Ω, R3 = 4Ω) in parallel, what is the total current supplied by the battery?
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Solution
Equivalent resistance R_eq = 1/(1/R1 + 1/R2 + 1/R3) = 1/(1/4 + 1/4 + 1/4) = 1.33Ω. Total current I = V/R_eq = 24V/1.33Ω = 18A.
Correct Answer: D — 6A
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