Electrostatics & Circuits Study for Competitive Exams

Electrostatics & Circuits

Coulombs Law and Electric Field Coulombs Law and Electric Field - Capacitance and Dielectrics Coulombs Law and Electric Field - Circuit Analysis Techniques Coulombs Law and Electric Field - Electric Field Problems Coulombs Law and Electric Field - Transient Response in RC Circuits Current, Resistance and Ohms Law Current, Resistance and Ohms Law - Capacitance and Dielectrics Current, Resistance and Ohms Law - Circuit Analysis Techniques Current, Resistance and Ohms Law - Electric Field Problems Current, Resistance and Ohms Law - Transient Response in RC Circuits DC Circuits and Kirchhoffs Laws DC Circuits and Kirchhoffs Laws - Capacitance and Dielectrics DC Circuits and Kirchhoffs Laws - Circuit Analysis Techniques DC Circuits and Kirchhoffs Laws - Electric Field Problems DC Circuits and Kirchhoffs Laws - Transient Response in RC Circuits Electric Potential and Capacitance Electric Potential and Capacitance - Capacitance and Dielectrics Electric Potential and Capacitance - Circuit Analysis Techniques Electric Potential and Capacitance - Electric Field Problems Electric Potential and Capacitance - Transient Response in RC Circuits Magnetic Fields and Electromagnetic Induction Magnetic Fields and Electromagnetic Induction - Capacitance and Dielectrics Magnetic Fields and Electromagnetic Induction - Circuit Analysis Techniques Magnetic Fields and Electromagnetic Induction - Electric Field Problems Magnetic Fields and Electromagnetic Induction - Transient Response in RC Circuits

Q. If two resistors, R1 and R2, are in series, what is the equivalent resistance (R_eq)?

A. R_eq = R1 + R2
B. R_eq = R1 * R2
C. R_eq = R1 / R2
D. R_eq = R1 - R2

Solution

The equivalent resistance of resistors in series is given by R_eq = R1 + R2.

Correct Answer: A — R_eq = R1 + R2

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Q. In a circuit with a 12 V battery and a 4 Ω resistor, what is the current flowing through the circuit?

A. 3 A
B. 2 A
C. 4 A
D. 1 A

Solution

Using Ohm's law, I = V/R = 12 V / 4 Ω = 3 A.

Correct Answer: B — 2 A

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Q. In a circuit with a 12 V battery and a resistor of 4 Ω, what is the current flowing through the circuit?

A. 3 A
B. 4 A
C. 2 A
D. 6 A

Solution

Using Ohm's Law: I = V/R = 12 V / 4 Ω = 3 A.

Correct Answer: A — 3 A

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Q. In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?

A. 4V
B. 6V
C. 8V
D. 12V

Solution

Total resistance = 4Ω + 8Ω = 12Ω. Current I = V / R = 12V / 12Ω = 1A. Voltage drop across 8Ω = I * R = 1A * 8Ω = 8V.

Correct Answer: B — 6V

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Q. In a circuit with a 24V battery and a total resistance of 6Ω, what is the total current flowing through the circuit?

A. 4A
B. 6A
C. 2A
D. 12A

Solution

Using Ohm's Law, I = V / R = 24V / 6Ω = 4A.

Correct Answer: A — 4A

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Q. In a circuit with a 24V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?

A. 8V
B. 12V
C. 16V
D. 20V

Solution

Total resistance R = 4Ω + 8Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop across 8Ω = I * R = 2A * 8Ω = 16V.

Correct Answer: B — 12V

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Q. In a circuit with a 24V battery and two resistors in series (8Ω and 4Ω), what is the voltage drop across the 4Ω resistor?

A. 6 V
B. 8 V
C. 12 V
D. 4 V

Solution

Total resistance = 8Ω + 4Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop across 4Ω = I * R = 2A * 4Ω = 8 V.

Correct Answer: A — 6 V

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Q. In a circuit with a 24V power supply and two resistors of 8Ω and 4Ω in series, what is the voltage drop across the 4Ω resistor?

A. 4.8 V
B. 8 V
C. 12 V
D. 16 V

Solution

Total resistance R = 8Ω + 4Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop V_R = I * R = 2A * 4Ω = 8 V.

Correct Answer: C — 12 V

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Q. In a discharging RC circuit, what is the expression for the voltage across the capacitor at time t?

A. V = V0*e^(-t/RC)
B. V = V0*(1 - e^(-t/RC))
C. V = V0*t
D. V = V0/t

Solution

In a discharging RC circuit, the voltage across the capacitor at time t is given by V = V0*e^(-t/RC), where V0 is the initial voltage.

Correct Answer: A — V = V0*e^(-t/RC)

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Q. In a parallel circuit with a 12 V battery and two resistors of 4 Ω and 6 Ω, what is the total current supplied by the battery?

A. 2 A
B. 3 A
C. 4 A
D. 1 A

Solution

Total resistance, R_eq = 1/(1/4 + 1/6) = 2.4 Ω. Total current, I = V/R_eq = 12 V / 2.4 Ω = 5 A.

Correct Answer: B — 3 A

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Q. In a parallel circuit with a 12 V battery, what is the voltage across each branch?

A. 12 V
B. 6 V
C. 24 V
D. 0 V

Solution

In a parallel circuit, the voltage across each branch is equal to the source voltage, which is 12 V.

Correct Answer: A — 12 V

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Q. In a parallel circuit with a 12 V supply, what is the voltage across each resistor?

A. 12 V
B. 6 V
C. 24 V
D. 0 V

Solution

In a parallel circuit, the voltage across each component is equal to the supply voltage, so it is 12 V.

Correct Answer: A — 12 V

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Q. In a parallel circuit with a 12V battery and two resistors (4Ω and 12Ω), what is the total current supplied by the battery?

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Solution

Total current I = V/R_total. R_total = 1/(1/4 + 1/12) = 3Ω. I = 12V / 3Ω = 4 A.

Correct Answer: C — 3 A

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Q. In a parallel circuit with a total voltage of 12 V, what is the voltage across each resistor?

A. 6 V
B. 12 V
C. 24 V
D. 0 V

Solution

In a parallel circuit, the voltage across each resistor is equal to the total voltage, so it is 12 V.

Correct Answer: B — 12 V

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Q. In a parallel circuit with three resistors of 2Ω, 3Ω, and 6Ω, what is the total resistance?

A. 1Ω
B. 2Ω
C. 3Ω
D. 4Ω

Solution

1/Rtotal = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6 = 1Ω.

Correct Answer: A — 1Ω

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Q. In a parallel circuit with two resistors of 6 ohms and 12 ohms, what is the equivalent resistance?

A. 4 ohms
B. 8 ohms
C. 2 ohms
D. 3 ohms

Solution

For parallel resistors, 1/R_eq = 1/R1 + 1/R2. Here, 1/R_eq = 1/6 + 1/12 = 2/12 + 1/12 = 3/12, so R_eq = 4 ohms.

Correct Answer: A — 4 ohms

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Q. In a parallel circuit with two resistors of 6Ω and 12Ω, what is the total resistance?

A. 4Ω
B. 2Ω
C. 3Ω
D. 1Ω

Solution

1/R_eq = 1/R1 + 1/R2 = 1/6 + 1/12 = 1/4. Therefore, R_eq = 4Ω.

Correct Answer: A — 4Ω

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Q. In a parallel RC circuit, what happens to the total current as the capacitor charges?

A. It increases.
B. It decreases.
C. It remains constant.
D. It oscillates.

Solution

As the capacitor charges, the current through the capacitor decreases, leading to a decrease in total current.

Correct Answer: B — It decreases.

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Q. In a parallel RC circuit, what happens to the voltage across the capacitor as time approaches infinity?

A. It approaches zero
B. It approaches the supply voltage
C. It oscillates
D. It becomes negative

Solution

In a parallel RC circuit, as time approaches infinity, the voltage across the capacitor approaches the supply voltage.

Correct Answer: B — It approaches the supply voltage

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Q. In a series circuit with a 10 Ω and a 20 Ω resistor, what is the total resistance?

A. 10 Ω
B. 20 Ω
C. 30 Ω
D. 5 Ω

Solution

Total resistance in series is R_total = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.

Correct Answer: C — 30 Ω

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Q. In a series circuit with a 12 V battery and two resistors of 3 Ω and 6 Ω, what is the current flowing through the circuit?

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Solution

Total resistance R_total = R1 + R2 = 3 Ω + 6 Ω = 9 Ω. Current I = V/R_total = 12 V / 9 Ω = 1.33 A, approximately 1 A.

Correct Answer: B — 2 A

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Q. In a series circuit with a 12 V battery and two resistors of 4 Ω and 6 Ω, what is the voltage across the 4 Ω resistor?

A. 4.8 V
B. 6 V
C. 8 V
D. 3.6 V

Solution

Total resistance R = 4 Ω + 6 Ω = 10 Ω. Current I = V/R = 12 V / 10 Ω = 1.2 A. Voltage across 4 Ω = I * R = 1.2 A * 4 Ω = 4.8 V.

Correct Answer: A — 4.8 V

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Q. In a series circuit with a 12 V battery and two resistors of 4 Ω and 6 Ω, what is the current flowing through the circuit?

A. 1.2 A
B. 0.8 A
C. 2 A
D. 1 A

Solution

Total resistance R = R1 + R2 = 4 Ω + 6 Ω = 10 Ω. Current I = V/R = 12 V / 10 Ω = 1.2 A.

Correct Answer: A — 1.2 A

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Q. In a series circuit with a 12V battery and two resistors (4Ω and 6Ω), what is the total current flowing through the circuit?

A. 0.5 A
B. 1 A
C. 1.2 A
D. 2 A

Solution

Total resistance R = R1 + R2 = 4Ω + 6Ω = 10Ω. Current I = V/R = 12V / 10Ω = 1.2 A.

Correct Answer: B — 1 A

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Q. In a series circuit with a 12V battery and two resistors of 4Ω and 6Ω, what is the total current flowing through the circuit?

A. 1A
B. 2A
C. 3A
D. 4A

Solution

The total resistance in series is R_total = R1 + R2 = 4Ω + 6Ω = 10Ω. Using Ohm's law, I = V/R_total = 12V / 10Ω = 1.2A.

Correct Answer: B — 2A

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Q. In a series circuit with two capacitors of capacitance 3 µF and 6 µF, what is the total capacitance?

A. 2 µF
B. 1 µF
C. 9 µF
D. 4 µF

Solution

Total capacitance in series is given by 1/C_total = 1/C1 + 1/C2. Thus, 1/C_total = 1/3 + 1/6 = 1/2, so C_total = 2 µF.

Correct Answer: A — 2 µF

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Q. In a series circuit with two resistors R1 and R2, what is the total resistance (R_total)?

A. R_total = R1 + R2
B. R_total = R1 * R2 / (R1 + R2)
C. R_total = R1 - R2
D. R_total = R1 / R2

Solution

In a series circuit, the total resistance is simply the sum of the individual resistances: R_total = R1 + R2.

Correct Answer: A — R_total = R1 + R2

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Q. In a series RC circuit, how does the total impedance (Z) relate to resistance (R) and reactance (Xc)?

A. Z = R + Xc
B. Z = R - Xc
C. Z = R*Xc
D. Z = R/Xc

Solution

In a series RC circuit, the total impedance is given by Z = R + Xc, where Xc is the capacitive reactance.

Correct Answer: A — Z = R + Xc

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Q. In a series RC circuit, if the resistance is doubled, how does the time constant change?

A. It remains the same.
B. It doubles.
C. It halves.
D. It quadruples.

Solution

The time constant τ = R * C, so if R is doubled, τ also doubles.

Correct Answer: B — It doubles.

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Q. In a series RC circuit, if the resistance is doubled, what happens to the time constant?

A. It remains the same.
B. It doubles.
C. It halves.
D. It quadruples.

Solution

The time constant τ = R * C will double if the resistance R is doubled, assuming capacitance C remains constant.

Correct Answer: B — It doubles.

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Electrostatics & Circuits

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