Current Electricity – II FiniteQuest

Current Electricity – II

Charge, Current, Potential, Potential difference and Resistance: Ohm’s law

1 / 10

Two resistors, R₁ = 4 Ω and R₂ = 6 Ω, are connected in parallel. What is the equivalent resistance of the combination?

10 Ω

2.4 Ω

4 Ω

1.5 Ω

Equivalent Resistance of Parallel Resistors

Physics Problem: Equivalent Resistance of Parallel Resistors

Two resistors, R₁ = 4 Ω and R₂ = 6 Ω, are connected in parallel. What is the equivalent resistance of the combination?

Given:

Resistor R₁ = 4 Ω
Resistor R₂ = 6 Ω

Formula:

The equivalent resistance (R_eq) of two resistors R₁ and R₂ connected in parallel is given by:

$Equivalent Resistance Formula$

To find R_eq, take the reciprocal of the sum of the reciprocals:

$Equivalent Resistance Calculation$

Calculation:

Substitute the given values into the formula:

$Substituted Values$

Find a common denominator and add the fractions:

$Common Denominator$

Now take the reciprocal of the result:

$Final Result$

Answer: The equivalent resistance of the combination when two resistors 4 Ω and 6 Ω are connected in parallel is 2.4 Ω.

Equivalent Resistance of Parallel Resistors

Physics Problem: Equivalent Resistance of Parallel Resistors

Two resistors, R₁ = 4 Ω and R₂ = 6 Ω, are connected in parallel. What is the equivalent resistance of the combination?

Given:

Resistor R₁ = 4 Ω
Resistor R₂ = 6 Ω

Formula:

The equivalent resistance (R_eq) of two resistors R₁ and R₂ connected in parallel is given by:

$Equivalent Resistance Formula$

To find R_eq, take the reciprocal of the sum of the reciprocals:

$Equivalent Resistance Calculation$

Calculation:

Substitute the given values into the formula:

$Substituted Values$

Find a common denominator and add the fractions:

$Common Denominator$

Now take the reciprocal of the result:

$Final Result$

Answer: The equivalent resistance of the combination when two resistors 4 Ω and 6 Ω are connected in parallel is 2.4 Ω.

2 / 10

If the potential difference across a conductor is tripled while the resistance remains constant, what happens to the current?

It remains the same

It is tripled

It is halved

It is reduced to one-third

3 / 10

In a series circuit with resistors R₁, R₂, and R₃, if R₁ = 2 Ω, R₂ = 3 Ω, and R₃ = 5 Ω, and the total voltage is 10 V, what is the current flowing through the circuit?

10 A

2 A

0.5 A

1 A

Given:

Resistor R₁ = 2 Ω
Resistor R₂ = 3 Ω
Resistor R₃ = 5 Ω
Total voltage, V_total = 10 V

Formula:

The total resistance (R_total) in a series circuit is the sum of the individual resistances:

R_total = R₁ + R₂ + R₃

The current (I) flowing through the circuit can be found using Ohm’s Law:

I = $\frac{V_{\text{total}}}{R_{\text{total}}}$

Calculation:

First, calculate the total resistance:

R_total = 2 Ω + 3 Ω + 5 Ω
R_total = 10 Ω

Next, use Ohm’s Law to find the current:

I = $\frac{10 \, \text{V}}{10 \, \Omega}$
I = 1 A

Answer: The current flowing through the circuit is 1 A.

Given:

Resistor R₁ = 2 Ω
Resistor R₂ = 3 Ω
Resistor R₃ = 5 Ω
Total voltage, V_total = 10 V

Formula:

The total resistance (R_total) in a series circuit is the sum of the individual resistances:

R_total = R₁ + R₂ + R₃

The current (I) flowing through the circuit can be found using Ohm’s Law:

I = $\frac{V_{\text{total}}}{R_{\text{total}}}$

Calculation:

First, calculate the total resistance:

R_total = 2 Ω + 3 Ω + 5 Ω
R_total = 10 Ω

Next, use Ohm’s Law to find the current:

I = $\frac{10 \, \text{V}}{10 \, \Omega}$
I = 1 A

Answer: The current flowing through the circuit is 1 A.

4 / 10

A 60 W light bulb operates on a 120 V power supply. What is the resistance of the light bulb?

240 Ω

60 Ω

2 Ω

120 Ω

Resistance of a Light Bulb

Physics Problem: Resistance of a Light Bulb

A 60 W light bulb operates on a 120 V power supply. What is the resistance of the light bulb?

Given:

Power, P = 60 W
Voltage, V = 120 V

Formula:

The resistance (R) of the light bulb can be calculated using the formula:

P = $\frac{V^2}{R}$

Rearrange the formula to solve for R:

R = $\frac{V^2}{P}$

Calculation:

Substitute the given values into the formula:

R = $\frac{(120 \, \text{V})^2}{60 \, \text{W}}$

Simplify the expression:

R = $\frac{14400 \, \text{V}^2}{60 \, \text{W}}$
R = 240 Ω

Answer: The resistance of the light bulb is 240 Ω.

Resistance of a Light Bulb

Physics Problem: Resistance of a Light Bulb

A 60 W light bulb operates on a 120 V power supply. What is the resistance of the light bulb?

Given:

Power, P = 60 W
Voltage, V = 120 V

Formula:

The resistance (R) of the light bulb can be calculated using the formula:

P = $\frac{V^2}{R}$

Rearrange the formula to solve for R:

R = $\frac{V^2}{P}$

Calculation:

Substitute the given values into the formula:

R = $\frac{(120 \, \text{V})^2}{60 \, \text{W}}$

Simplify the expression:

R = $\frac{14400 \, \text{V}^2}{60 \, \text{W}}$
R = 240 Ω

Answer: The resistance of the light bulb is 240 Ω.

5 / 10

The resistance of a wire changes with temperature. If the temperature coefficient of resistance for a material is 0.004 , what is the percentage change in resistance when the temperature increases from 20°C to 120°C?

40%

0.4%

400%

Percentage Change in Resistance with Temperature

Physics Problem: Temperature Dependence of Resistance

The resistance of a wire changes with temperature. If the temperature coefficient of resistance for a material is 0.004 °C^-1, what is the percentage change in resistance when the temperature increases from 20°C to 120°C?

Given:

Temperature coefficient of resistance, α = 0.004 °C^-1
Initial temperature, T₀ = 20°C
Final temperature, T_f = 120°C

Formula:

The resistance at temperature t is given by:

R_t = R₀ (1 + α ΔT)

Where ΔT is the change in temperature:

ΔT = T_f – T₀

Calculation:

Calculate the change in temperature:

ΔT = 120°C – 20°C = 100°C

Substitute the values into the formula:

R_t = R₀ (1 + 0.004 × 100)

Simplify the expression:

R_t = R₀ (1 + 0.4)
R_t = R₀ × 1.4

The percentage change in resistance is:

(1.4 – 1) × 100% = 0.4 × 100% = 40%

Answer: The percentage change in resistance when the temperature increases from 20°C to 120°C is 40%.

Percentage Change in Resistance with Temperature

Physics Problem: Temperature Dependence of Resistance

Given:

Temperature coefficient of resistance, α = 0.004 °C^-1
Initial temperature, T₀ = 20°C
Final temperature, T_f = 120°C

Formula:

The resistance at temperature t is given by:

R_t = R₀ (1 + α ΔT)

Where ΔT is the change in temperature:

ΔT = T_f – T₀

Calculation:

Calculate the change in temperature:

ΔT = 120°C – 20°C = 100°C

Substitute the values into the formula:

R_t = R₀ (1 + 0.004 × 100)

Simplify the expression:

R_t = R₀ (1 + 0.4)
R_t = R₀ × 1.4

The percentage change in resistance is:

(1.4 – 1) × 100% = 0.4 × 100% = 40%

Answer: The percentage change in resistance when the temperature increases from 20°C to 120°C is 40%.

6 / 10

If a resistor is labeled with the color code brown, black, red, and gold, what is its resistance and tolerance?

100 Ω ±5%

10 Ω ±10%

1 kΩ ±5%

10 kΩ ±10%

Charge Stored on a Capacitor

Physics Problem: Charge Stored on a Capacitor

A 12 V battery is connected to a parallel plate capacitor with a capacitance of 100 μF. What is the charge stored on the capacitor?

Given:

Voltage, V = 12 V
Capacitance, C = 100 μF = 100 × 10^-6 F

Formula:

The charge (Q) stored on a capacitor is given by:

Q = C × V

Calculation:

Substitute the given values into the formula:

Q = (100 × 10^-6 F) × 12 V

Simplify the expression:

Q = 1.2 × 10^-3 C = 1.2 mC

Answer: The charge stored on the capacitor is 1.2 mC (milliCoulombs).

Charge Stored on a Capacitor

Physics Problem: Charge Stored on a Capacitor

A 12 V battery is connected to a parallel plate capacitor with a capacitance of 100 μF. What is the charge stored on the capacitor?

Given:

Voltage, V = 12 V
Capacitance, C = 100 μF = 100 × 10^-6 F

Formula:

The charge (Q) stored on a capacitor is given by:

Q = C × V

Calculation:

Substitute the given values into the formula:

Q = (100 × 10^-6 F) × 12 V

Simplify the expression:

Q = 1.2 × 10^-3 C = 1.2 mC

Answer: The charge stored on the capacitor is 1.2 mC (milliCoulombs).

7 / 10

A 12 V battery is connected to a parallel plate capacitor with a capacitance of 100 μF. What is the charge stored on the capacitor?

1.2 mC

12 mC

120 mC

1.2 C

Physics Problem: Charge Stored on a Capacitor

A 12 V battery is connected to a parallel plate capacitor with a capacitance of 100 μF. What is the charge stored on the capacitor?

Given:

Voltage, V = 12 V
Capacitance, C = 100 μF = 100 × 10^-6 F

Formula:

The charge (Q) stored on a capacitor is given by:

Q = C × V

Calculation:

Substitute the given values into the formula:

Q = (100 × 10^-6 F) × 12 V

Simplify the expression:

Q = 1.2 × 10^-3 C = 1.2 mC

Answer: The charge stored on the capacitor is 1.2 mC (milliCoulombs).

Physics Problem: Charge Stored on a Capacitor

A 12 V battery is connected to a parallel plate capacitor with a capacitance of 100 μF. What is the charge stored on the capacitor?

Given:

Voltage, V = 12 V
Capacitance, C = 100 μF = 100 × 10^-6 F

Formula:

The charge (Q) stored on a capacitor is given by:

Q = C × V

Calculation:

Substitute the given values into the formula:

Q = (100 × 10^-6 F) × 12 V

Simplify the expression:

Q = 1.2 × 10^-3 C = 1.2 mC

Answer: The charge stored on the capacitor is 1.2 mC (milliCoulombs).

8 / 10

A point charge of 2 μC is located in a vacuum. What is the electric potential at a point 3 meters away from the charge?
(Given

k = 2 μC = 2 × 10⁹ N m^{2 /C²}

)

6 × 10³ V

5.99 × 10² V

5.99 × 10³ V

6 × 10² V

Electric Potential Calculation

Physics Problem: Electric Potential

A point charge of 2 μC is located in a vacuum. What is the electric potential at a point 3 meters away from the charge?

Given:

Charge, Q = 2 μC = 2 × 10^-6 C
Distance, r = 3 m
Coulomb’s constant, k = 8.99 × 10⁹ N m²/C²

Formula:

The electric potential (V) at a distance (r) from a point charge (Q) is given by:

$Electric Potential Formula$

Calculation:

Substitute the given values into the formula:

$Electric Potential Calculation$

Simplify the expression:

$Simplified Electric Potential Calculation$

$Further Simplified Electric Potential Calculation$

$Final Electric Potential$

Answer: The electric potential at a point 3 meters away from the charge is 5.99 × 10³ V.

Electric Potential Calculation

Physics Problem: Electric Potential

A point charge of 2 μC is located in a vacuum. What is the electric potential at a point 3 meters away from the charge?

Given:

Charge, Q = 2 μC = 2 × 10^-6 C
Distance, r = 3 m
Coulomb’s constant, k = 8.99 × 10⁹ N m²/C²

Formula:

The electric potential (V) at a distance (r) from a point charge (Q) is given by:

$Electric Potential Formula$

Calculation:

Substitute the given values into the formula:

$Electric Potential Calculation$

Simplify the expression:

$Simplified Electric Potential Calculation$

$Further Simplified Electric Potential Calculation$

$Final Electric Potential$

Answer: The electric potential at a point 3 meters away from the charge is 5.99 × 10³ V.

9 / 10

Two charges, Q1 and Q2, are placed at a distance r apart. If the magnitude of each charge is doubled and the distance between them is halved, the force between them will be:

Quadrupled

Halved

Doubled

Increased by sixteen times

Coulomb’s Law Problem

Physics Problem: Coulomb’s Law

Two charges, Q₁ and Q₂, are placed at a distance r apart. If the magnitude of each charge is doubled and the distance between them is halved, the force between them will be:

Options:

A. Quadrupled
B. Halved
C. Doubled
D. Increased by sixteen times

Answer: D. Increased by sixteen times

Explanation:

The force between two charges is given by Coulomb’s law:

$Coulomb's Law$

When the magnitude of each charge is doubled, the charges become 2Q₁ and 2Q₂. When the distance is halved, the distance becomes r/2.

Substituting these new values into Coulomb’s law:

$Modified Coulomb's Law$

Simplifying the expression:

$Simplified Coulomb's Law$

$Final Simplified Coulomb's Law$

The original force F is given by:

$Original Coulomb's Law$

Therefore, the new force F’ is:

$New Force$

Thus, the force between the charges will be increased by sixteen times.

Coulomb’s Law Problem

Physics Problem: Coulomb’s Law

Two charges, Q₁ and Q₂, are placed at a distance r apart. If the magnitude of each charge is doubled and the distance between them is halved, the force between them will be:

Options:

A. Quadrupled
B. Halved
C. Doubled
D. Increased by sixteen times

Answer: D. Increased by sixteen times

Explanation:

The force between two charges is given by Coulomb’s law:

$Coulomb's Law$

When the magnitude of each charge is doubled, the charges become 2Q₁ and 2Q₂. When the distance is halved, the distance becomes r/2.

Substituting these new values into Coulomb’s law:

$Modified Coulomb's Law$

Simplifying the expression:

$Simplified Coulomb's Law$

$Final Simplified Coulomb's Law$

The original force F is given by:

$Original Coulomb's Law$

Therefore, the new force F’ is:

$New Force$

Thus, the force between the charges will be increased by sixteen times.

10 / 10

A current of 5 A flows through a conductor for 2 minutes. What is the total charge that has passed through the conductor?

1200 C

10 C

300 C

600 C

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