Physics Study Guide: Electricity and Circuits

Q14. A 1200 W device runs on 120 V. What is the current?

Background

Topic: Electric Power and Current

This question tests your understanding of the relationship between power, voltage, and current in electrical circuits.

Key formula:

$P = IV$

Where:

• $P$ = power (in watts, W)

• $I$ = current (in amperes, A)

• $V$ = voltage (in volts, V)

Step-by-Step Guidance

1. Identify the known values: $P = 1200\ \text{W}$, $V = 120\ \text{V}$.

2. Write the formula relating power, current, and voltage: $P = IV$.

3. Rearrange the formula to solve for current: $I = \frac{P}{V}$.

4. Set up the calculation by plugging in the values, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Q15. Two resistors (6 Ω and 12 Ω) are in parallel across 12 V. What is the current through the 12 Ω resistor?

Background

Topic: Parallel Circuits and Ohm's Law

This question tests your ability to analyze parallel resistor circuits and apply Ohm's Law to find the current through a specific resistor.

Key formula:

$I = \frac{V}{R}$

Where:

• $I$ = current through the resistor (A)

• $V$ = voltage across the resistor (V)

• $R$ = resistance (Ω)

Step-by-Step Guidance

1. Recognize that in parallel circuits, each resistor has the same voltage across it: $V = 12\ \text{V}$.

2. Focus on the 12 Ω resistor. Use Ohm's Law: $I = \frac{V}{R}$.

3. Set up the calculation for the current through the 12 Ω resistor: $I = \frac{12}{12}$.

4. Do not compute the final value yet; leave the setup for the student to finish.

Try solving on your own before revealing the answer!

Q16. Adding more resistors in parallel makes the equivalent resistance:

Background

Topic: Parallel Circuits and Equivalent Resistance

This question tests your understanding of how equivalent resistance changes when more resistors are added in parallel.

Key formula:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

Where:

• $R_{eq}$ = equivalent resistance

• $R_1, R_2, \ldots$ = individual resistances

Step-by-Step Guidance

1. Recall the formula for equivalent resistance in parallel circuits.

2. Think about what happens to $R_{eq}$ as you add more terms to the denominator.

3. Consider the effect of adding a resistor with finite resistance to the parallel network.

Try solving on your own before revealing the answer!

Q17. If resistance doubles and capacitance halves, what happens to the time constant?

Background

Topic: RC Circuits and Time Constant

This question tests your understanding of the time constant in RC circuits and how it changes with resistance and capacitance.

Key formula:

$\tau = RC$

Where:

• $\tau$ = time constant (seconds)

• $R$ = resistance (Ω)

• $C$ = capacitance (F)

Step-by-Step Guidance

1. Write the original formula for the time constant: $\tau = RC$.

2. Consider what happens if $R$ is doubled ($2R$) and $C$ is halved ($\frac{C}{2}$).

3. Set up the new time constant: $\tau' = (2R) \times \left(\frac{C}{2}\right)$.

4. Simplify the expression, but stop before stating the final result.

Try solving on your own before revealing the answer!

Q18. A heater uses 18 kW at 240 V. Find the resistance.

Background

Topic: Electric Power and Resistance

This question tests your ability to relate power, voltage, and resistance in electrical devices.

Key formula:

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

Where:

• $P$ = power (W)

• $V$ = voltage (V)

• $R$ = resistance (Ω)

Step-by-Step Guidance

1. Identify the known values: $P = 18,000\ \text{W}$, $V = 240\ \text{V}$.

2. Write the formula relating power, voltage, and resistance: $P = \frac{V^2}{R}$.

3. Rearrange the formula to solve for resistance: $R = \frac{V^2}{P}$.

4. Set up the calculation by plugging in the values, but do not compute the final answer yet.

Try solving on your own before revealing the answer!

Q19. Inserting a dielectric into a capacitor:

Background

Topic: Capacitance and Dielectrics

This question tests your understanding of how a dielectric affects the capacitance of a capacitor.

Key formula:

$C = \kappa \varepsilon_0 \frac{A}{d}$

Where:

• $C$ = capacitance (F)

• $\kappa$ = dielectric constant (dimensionless)

• $\varepsilon_0$ = permittivity of free space

• $A$ = area of plates (m²)

• $d$ = separation between plates (m)

Step-by-Step Guidance

1. Recall the formula for capacitance with a dielectric.

2. Consider what happens to $C$ when $\kappa$ increases (inserting a dielectric).

3. Think about the physical meaning: does capacitance increase, decrease, or stay the same?

Try solving on your own before revealing the answer!

Q20. A 500 W device runs 6 hr/day. Electricity costs $0.12/kWh. What is the annual cost?

Background

Topic: Energy Consumption and Cost Calculation

This question tests your ability to calculate energy usage and cost over time.

Key formula:

$\text{Energy (kWh)} = \frac{\text{Power (W)} \times \text{Time (h)}}{1000}$

$\text{Annual Cost} = \text{Energy (kWh)} \times \text{Rate (\$/kWh)}$

Step-by-Step Guidance

1. Calculate daily energy usage: $\text{Energy} = \frac{500 \times 6}{1000}$.

2. Multiply by 365 to get annual energy usage.

3. Multiply annual energy usage by the cost per kWh: $0.12$.

4. Set up the calculation, but stop before computing the final cost.

Try solving on your own before revealing the answer!