Problem 70 - Physics for Scientists and Engineers with Modern Physics

Question

Suppose that a person's body resistance is 950\(\Omega\) (a) What current passes through the body when the person accidentally is connected to 110 \(\mathrm{V} ?(b)\) If there is an alternative path to ground whose resistance is \(35 \Omega,\) what current passes through the person? \((c)\) If the voltage source can produce at most 1.5 \(\mathrm{A}\) , how much current passes through the person in case \((b) ?\)

Step-by-Step Solution

Answer

(a) 0.116 A, (b) 0.12 A, (c) 0.054 A

Step 1: Understanding Ohm's Law

Ohm's law is defined as \( V = I \times R \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. We'll use this law to calculate the current passing through the body in different scenarios.

Step 2: Calculate Current Through Body With Direct Connection

For part (a), we're given a body resistance \( R = 950\, \Omega \) and a voltage \( V = 110\, V \). Using Ohm's law, solve for current \( I \) as follows: \[ I = \frac{V}{R} = \frac{110}{950} \approx 0.116\, A \] So, the current passing through the body is approximately 0.116 A.

Step 3: Calculate Parallel Resistance

For part (b), we need to consider an alternative path to ground with resistance \( 35\, \Omega \), creating a parallel circuit. The equivalent resistance \( R_{eq} \) can be calculated using the formula for parallel resistances: \[ \frac{1}{R_{eq}} = \frac{1}{950} + \frac{1}{35} \] Solving this gives: \[ R_{eq} \approx \frac{1}{0.00105} \approx 33.33\, \Omega \]

Step 4: Calculate Current Through Body With Parallel Path

With the equivalent resistance, we calculate the total current through the body and path using Ohm’s law again. For \( V = 110\, V \) and \( R_{eq} = 33.33\, \Omega \): \[ I_{total} = \frac{110}{33.33} \approx 3.3\, A \] Next, find current through the body using the current division rule: \[ I_{body} = I_{total} \times \frac{R_{alternate}}{R_{body} + R_{alternate}} = 3.3 \times \frac{35}{950 + 35} \approx 0.12\, A \] So, approximately 0.12 A flows through the person.

Step 5: Current Limited by Voltage Source

For part (c), since the max current from the source is 1.5 A, we must recalculate the current through the body with this limitation. Using the current division principle again for total \( I = 1.5\, A \): \[ I_{body} = 1.5 \times \frac{35}{950 + 35} \approx 0.054\, A \] Hence, approximately 54 mA passes through the person in this scenario.

Key Concepts

Body Resistance

Body resistance refers to how a person's body resists the flow of electric current. This resistance is significant because it affects the amount of current that flows through the body when it encounters a voltage source. Normally, the human body's internal resistance ranges from approximately 500 ohms to 1000 ohms, depending on various factors such as skin condition and internal body conductivity.
Understanding this concept is crucial because higher resistance means less current flows through the body, which is safer. In our exercise, the body resistance is specified as 950 ohms.

• This value affects the outcome when calculating the current using Ohm's Law, which states that current (\( I \)) equals the voltage (\( V \)) divided by the resistance (\( R \)).
• A person with higher body resistance would experience a lower electrical current for the same voltage applied.

Being aware of the body resistance allows us to anticipate potential risks when exposed to electrical circuits, ensuring safer handling of electrical equipment.

Parallel Circuit

In electrical terms, a parallel circuit is one where the components are connected alongside each other, providing multiple pathways for the current to flow. This kind of connection impacts the overall resistance of the circuit, as adding more paths usually reduces the total or equivalent resistance.
In our exercise, there was an additional path for the current with a resistance of 35 ohms, creating a parallel circuit between that path and the human body. To find the equivalent resistance (\( R_{eq} \)) in a parallel circuit, we apply the formula: \[\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} +\ldots \frac{1}{R_{n}}\]where\( R_1, R_2, \ldots, R_n \)are the individual resistances.

• For two resistors like in our scenario, the effective resistance is always less than the smallest individual resistance.
• This configuration ensures that any failure in one pathway doesn’t break the whole circuit.

Parallel circuits are integral in many electrical designs because they ensure that devices receive the same voltage, enabling them to function correctly without being affected by the failure of one component.

Current Division Rule

The current division rule is a technique used to calculate the current flowing through individual components in a parallel circuit. When you have multiple resistors connected in parallel, the total current divides itself among the parallel branches inversely in proportion to their resistances.
The relationship is given by the formula for any resistor \( R_n \) in a parallel circuit,\[I_n = I_{total} \times \frac{R_{total}}{R_n}\]where \( I_n \) is the current through the resistor, \( I_{total} \) is the total current in the circuit, \( R_{total} \) is the equivalent resistance, and \( R_n \) is the resistance of the resistor in question.

• In the exercise, we used this rule to calculate the current through the body when there's another path present.
• It helps in understanding the distribution of current, making it easier to design circuits and prevent overloads.

Understanding this rule is fundamental in circuit design and analysis as it ensures the efficient and safe flow of current across various components.