Problem 70 - University Physics with Modern Physics

Question

A person with body resistance between his hands of 10 \(\mathrm{k} \Omega\) accidentally grasps the terminals of a \(14-\mathrm{kV}\) power supply. (a) If the internal resistance of the power supply is \(2000 \Omega,\) what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 \(\mathrm{mA}\) or less?

Step-by-Step Solution

Answer

(a) 1.167 A; (b) 13625 W; (c) 13.99 MΩ.

Step 1: Understanding the Circuit

Consider a circuit consisting of a body resistance of \(10 \, \text{k}\Omega\) and an internal resistance of the power supply of \(2000 \, \Omega\). When the person touches the terminals, the body resistance and the power supply's internal resistance are in series.

Step 2: Calculate Total Resistance

For series resistors, the total resistance \(R_{\text{total}}\) is the sum of the body resistance \(R_b = 10000 \, \Omega\) and the internal resistance \(R_i = 2000 \, \Omega\). Thus:\[ R_{\text{total}} = R_b + R_i = 10000 \, \Omega + 2000 \, \Omega = 12000 \, \Omega \]

Step 3: Find the Current Through the Body

Using Ohm's law, \( V = IR \), where \( V = 14000 \, \text{V} \). Solving for the current \(I\), we get:\[ I = \frac{V}{R_{\text{total}}} = \frac{14000 \, \text{V}}{12000 \, \Omega} = 1.167 \, \text{A} \]

Step 4: Calculate the Power Dissipated in the Body

Power dissipated \(P\) can be calculated using \( P = I^2 R \). With \(I = 1.167 \, \text{A}\) and \(R_b = 10000 \, \Omega\):\[ P = (1.167 \, \text{A})^2 \times 10000 \, \Omega = 13625 \, \text{W} \]

Step 5: Determine Required Internal Resistance for Safety

To make the supply safe, the current must be less than \(1.00 \, \text{mA} = 0.001 \, \text{A}\). Using \(I = \frac{V}{R_{\text{total}}}\) and solving for \( R_{\text{total}} = \frac{V}{I} \), we have:\[ R_{\text{total}} = \frac{14000 \, \text{V}}{0.001 \, \text{A}} = 14000000 \, \Omega \]Since \(R_{\text{total}} = R_b + R_i\), we find:\[ R_i = R_{\text{total}} - R_b = 14000000 \, \Omega - 10000 \, \Omega = 13990000 \, \Omega \]

Key Concepts

Resistance Calculation

When it comes to understanding resistance calculation in a circuit, it's helpful to visualize how resistance works in series circuits. When resistors are in series, their resistances simply add up. This is because the current flowing through the circuit experiences each resistance one after the other. For instance, in our given scenario, the person's body has a resistance of \(10 \text{k}\Omega\), and the power supply has an internal resistance of \(2000\,\Omega\).

To find the total resistance, you add the two resistances together:

• Body Resistance, \( R_b = 10000 \, \Omega \)
• Internal Resistance, \( R_i = 2000 \, \Omega \)

Combining these gives a Total Resistance, \( R_{\text{total}} = R_b + R_i = 12000 \, \Omega \).

Understanding this concept helps you figure out the current flow through the circuit using Ohm's Law. The total resistance plays a crucial role in determining how much current will pass through various elements of the circuit.

Power Dissipation

Power dissipation in a circuit is another key concept, especially in safety considerations. Power dissipation refers to the conversion of electrical energy into heat within an element, like a resistor. Its calculation is essential to ensure that components do not overheat.

To calculate power dissipated, you can use the formula:
\[ P = I^2 \, R \]
Where:

• \(P\) is power in watts
• \(I\) is the current in amperes
• \(R\) is the resistance in ohms

In our example, with the current calculated at \(1.167 \, \text{A}\) through body resistance of \(10000 \, \Omega\), the power dissipated can be calculated as:
\[ P = (1.167 \, \text{A})^2 \times 10000 \, \Omega = 13625 \, \text{W} \]

Such a high power indicates a significant conversion of electrical energy into heat, which can be dangerous. Thus, knowing how to calculate power dissipation helps in designing circuits that are safe and reliable.

Series Circuit Analysis

Understanding series circuit analysis is fundamental for solving problems involving multiple electrical components. In a series circuit, all components are connected end-to-end, forming a single path for current flow. The unique characteristic of series circuits is that the same current flows through all components, but the voltage across each component can vary based on its resistance.

In practical terms:

• Adding more resistors increases the total resistance
• The total voltage across the circuit equals the sum of voltages across each resistor
• The overall current in the series circuit is determined by the total resistance

To ensure a circuit is safe, you must consider both the resistances present and the necessity to manage potential maximum current levels. For example, to make the power supply safe with a maximum current of \(1.00 \, \text{mA}\), internal resistance needs adjusting. With the total potential voltage at \(14000 \, \text{V}\), the calculation becomes:
\[ R_{\text{total}} = \frac{14000 \, \text{V}}{0.001 \, \text{A}} = 14000000 \, \Omega \]
Hence, to find the safer internal resistance:
\[ R_i = 14000000 \, \Omega - 10000 \, \Omega = 13990000 \, \Omega \]

Series circuit analysis helps understand how changes to one part of the circuit can affect the whole system, ensuring better design and safety considerations.