Answered

The electrical current, $I$ (amps), passing through an electronic component varies directly with the applied voltage, $V$ (volts), according to the relationship $V=I \cdot R$, where $R$ is the resistance measured in Ohms ( $\Omega$ ).



The resistance $R$ of a number of components wired in a series (one after the other) is simply the sum of all the resistances of the individual components.



Five identical electronic components were connected in series. A fixed but unknown voltage placed across them caused a 2.5 amp current to flow. When two of the components were replaced with standard $10\,\Omega$ resistors, the current dropped to 1.9 amps. What is the resistance of each component?

Verified Answer :

  • Analyze the first circuit scenario: \(V = 2.5 \times 5R_c = 12.5R_c\).
  • Analyze the second circuit scenario: \(V = 1.9 \times (3R_c + 20) = 5.7R_c + 38\).
  • Equate the voltages and solve for \(R_c\): \(12.5R_c = 5.7R_c + 38\).
  • The calculated resistance of each component is \(\boxed{\frac{95}{17} \, \Omega}\).

Explanation

  1. Analyzing the First Circuit Configuration

The problem describes an electrical circuit with five identical components connected in series. Let \(R_c\) be the resistance of each component in Ohms (\(\Omega\)). The total resistance of five components in series is the sum of their individual resistances: \(R_{total,1} = 5R_c\). The circuit is connected to a fixed, unknown voltage \(V\). According to Ohm's Law (\(V = I \times R\)), the relationship for the first scenario with a current \(I_1 = 2.5\) A is \(V = 2.5 \times (5R_c) = 12.5R_c\).

  1. Analyzing the Second Circuit Configuration

In the second scenario, two of the original components (each with resistance \(R_c\)) are replaced by two resistors, each with a resistance of \(10 \Omega\). The circuit now contains three components of resistance \(R_c\) and two resistors of resistance \(10 \Omega\), all in series. The total resistance of this configuration is \(R_{total,2} = 3R_c + 2 \times 10 = 3R_c + 20\). The voltage \(V\) remains the same as in the first scenario, and the new current is \(I_2 = 1.9\) A. Applying Ohm's Law again, we have \(V = 1.9 \times (3R_c + 20) = 5.7R_c + 38\).

  1. Solving for the Resistance \(R_c\)

Since the voltage \(V\) is the same in both scenarios, we can set the two expressions for \(V\) equal to each other: \(12.5R_c = 5.7R_c + 38\). To solve for \(R_c\), subtract \(5.7R_c\) from both sides of the equation: \(12.5R_c - 5.7R_c = 38\), which simplifies to \(6.8R_c = 38\). Dividing both sides by \(6.8\), we get \(R_c = \frac{38}{6.8}\). Performing this division, we find \(R_c \approx 5.588 \Omega\). More precisely, the fraction is \(\frac{380}{68} = \frac{95}{17} \Omega\).

  1. Final Conclusion

The resistance of each of the five identical electronic components is \(\frac{95}{17} \Omega\), which is approximately \(5.59 \Omega\). Thus, the exact resistance value is \(\boxed{\frac{95}{17} \Omega}\).

Examples

Understanding how resistance affects current in series circuits is fundamental in electrical engineering and DIY electronics. For example, if you are designing a circuit for an LED project and need to control the brightness, adding resistors in series is a common way to limit the current and protect your components. By calculating the required resistance, you can ensure that your devices operate within safe and optimal parameters, preventing damage from excessive current flow.

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