Problem 103 - Physics

Question

A fax machine uses 0.110 A of current in its normal mode of operation, but only 0.067 A in the standby mode. The machine uses a potential difference of \(120 \mathrm{~V}\). (a) In one minute, how much more charge passes through the machine in the normal mode versus the standby mode, and (b) how much more energy is used?

Step-by-Step Solution

Answer

(a) 2.58 C more charge; (b) 309.6 J more energy.

Step 1: Understanding the Problem

We need to find how much more charge and energy the fax machine uses in normal mode compared to standby mode over one minute. Current (I) is given in amperes, potential difference (V) is given in volts, and time is one minute (or 60 seconds).

Step 2: Calculate Charge in Normal and Standby Mode

Charge (Q) is calculated using the formula: \(Q = I \times t\). For normal mode, with \(I = 0.110\, \text{A}\) and \(t = 60\, \text{s}\), the charge \(Q_n = 0.110 \times 60\). For standby mode, with \(I = 0.067\, \text{A}\), the charge \(Q_s = 0.067 \times 60\).

Step 3: Calculate the Difference in Charge

Subtract the charge in standby mode from the charge in normal mode to find the additional charge: \(\Delta Q = Q_n - Q_s = (0.110 \times 60) - (0.067 \times 60)\).

Step 4: Calculate Energy in Normal and Standby Mode

Energy (E) is calculated using the formula: \(E = V \times Q\). For normal mode, \(E_n = 120 \times Q_n\). For standby mode, \(E_s = 120 \times Q_s\).

Step 5: Calculate the Difference in Energy

Subtract the energy in standby mode from the energy in normal mode to find the additional energy: \(\Delta E = E_n - E_s\).

Step 6: Final Calculations and Results

Calculate the exact values: \(Q_n = 6.6\, \text{C}\) and \(Q_s = 4.02\, \text{C}\), making \(\Delta Q = 2.58\, \text{C}\). \(E_n = 792\, \text{J}\) and \(E_s = 482.4\, \text{J}\), making \(\Delta E = 309.6\, \text{J}\).

Key Concepts

Current in Circuits

Electric current is the flow of electric charge through a circuit, usually carried by moving electrons in a wire. The unit of electric current is the ampere (A), which represents the flow of one coulomb of charge per second. In an electric circuit, the current is the same at any point in a series circuit. This is because the charge that enters a section of the circuit must equal the charge that leaves it.
In the context of the fax machine, the current used in the normal mode is 0.110 A, while in the standby mode, it is 0.067 A. Both currents represent the number of charges flowing per second, but the normal mode uses more current because it requires more power to operate.

• Current expressed as amperes (A)
• Determines the rate of charge flow
• Essential for understanding electrical consumption in devices

Understanding the current within a circuit helps us calculate other parameters such as charge and energy consumed over time, which are crucial for determining the efficiency and cost of operating electronic devices.

Potential Difference

Potential difference, also known as voltage, measures the difference in electric potential energy between two points in a circuit. It is a critical factor that drives the current through the circuit, as it provides the required force or push. The unit of potential difference is the volt (V).
In the fax machine example, a potential difference of 120 V is provided. This potential difference remains constant whether the machine is in normal or standby mode. The potential difference ensures that the electrical charges can move through the circuit efficiently, overcoming any resistance present.
Key characteristics of potential difference include:

• Expressed in volts (V)
• Essential for maintaining current flow
• A constant voltage is required for consistent device performance

By understanding potential difference, we can comprehend how various devices accomplish their unique tasks with distinct power requirements, ensuring they operate correctly and reliably.

Electrical Power Calculation

Electrical power is the rate at which electrical energy is consumed in a circuit. It is calculated by multiplying the potential difference (voltage) by the current. The unit of power is the watt (W), which represents one joule per second. Power can be calculated using the formula, \( P = V \times I \), where \( V \) is the potential difference and \( I \) is the current.
In this exercise, knowing the power consumption in both the normal and standby modes of a fax machine helps determine the energy usage difference. While the potential difference remains the same at 120 V, the differing currents in the normal (0.110 A) and standby (0.067 A) modes result in different power values. Here's how to calculate the power:

• Multiply voltage by current
• Normal mode: \( P = 120\, V \times 0.110\, A = 13.2\, W \)
• Standby mode: \( P = 120\, V \times 0.067\, A = 8.04\, W \)

By calculating power, we can deeply understand how much energy is used over time, which helps in managing electricity consumption and efficiency.