Calculating Current In Series Circuit With Resistors And Cells

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#article Two resistors with resistances of 55Ω and 75Ω connected in series are connected to four cells, each with a voltage of 1.5V and an internal resistance of 0.8Ω. The cells are connected in series. Our aim is to calculate the current flowing through this circuit. This exploration delves into the fundamental principles of electrical circuits, specifically focusing on series connections, Ohm's Law, and the impact of internal resistance within voltage sources.

Understanding Series Circuits

In electrical circuits, components can be connected in series or parallel. In a series circuit, components are connected end-to-end, forming a single path for the current to flow. This means that the current flowing through each component in a series circuit is the same. This fundamental characteristic is crucial for analyzing and calculating circuit behavior.

Resistors in Series

When resistors are connected in series, their resistances add up to give the total resistance of the circuit. This is because the current must flow through each resistor sequentially, experiencing a voltage drop across each one. The total resistance (Rtotal{R_\text{total}}) of resistors connected in series is calculated as:

Rtotal=R1+R2+R3+...{R_\text{total} = R_1 + R_2 + R_3 + ...}

In our case, we have two resistors with resistances of 55Ω and 75Ω connected in series. Therefore, the total resistance due to these resistors is:

Rresistors=55Ω+75Ω=130Ω{R_\text{resistors} = 55 \Omega + 75 \Omega = 130 \Omega}

Cells in Series

Similar to resistors, when cells (or batteries) are connected in series, their voltages add up. This is because the potential difference across each cell contributes to the overall potential difference of the series combination. The total voltage (Vtotal{V_\text{total}}) of cells connected in series is calculated as:

Vtotal=V1+V2+V3+...{V_\text{total} = V_1 + V_2 + V_3 + ...}

However, cells also have internal resistance, which must be considered. Internal resistance is the resistance within the cell itself, due to the materials and construction of the cell. When cells are connected in series, their internal resistances also add up, just like resistors. This total internal resistance reduces the actual voltage available to the external circuit.

In our scenario, we have four cells, each with a voltage of 1.5V and an internal resistance of 0.8Ω, connected in series. The total voltage provided by the cells is:

Vcells=4×1.5V=6V{V_\text{cells} = 4 \times 1.5 V = 6 V}

The total internal resistance of the cells is:

Rinternal=4×0.8Ω=3.2Ω{R_\text{internal} = 4 \times 0.8 \Omega = 3.2 \Omega}

Applying Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits that relates voltage (V), current (I), and resistance (R). It states that the current flowing through a conductor is directly proportional to the voltage across it and inversely proportional to its resistance. Mathematically, Ohm's Law is expressed as:

V=IR{V = IR}

To calculate the current flowing through our circuit, we need to consider the total resistance in the circuit, which includes both the external resistors and the internal resistance of the cells. The total resistance (Rtotal{R_\text{total}}) of the circuit is the sum of the resistance of the resistors and the internal resistance of the cells:

Rtotal=Rresistors+Rinternal=130Ω+3.2Ω=133.2Ω{R_\text{total} = R_\text{resistors} + R_\text{internal} = 130 \Omega + 3.2 \Omega = 133.2 \Omega}

Now, we can use Ohm's Law to calculate the current (I) flowing through the circuit. We know the total voltage (Vcells=6V{V_\text{cells} = 6 V}) and the total resistance (Rtotal=133.2Ω{R_\text{total} = 133.2 \Omega}). Rearranging Ohm's Law to solve for current, we get:

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

Substituting the values, we get:

I=6V133.2Ω≈0.045A{I = \frac{6 V}{133.2 \Omega} \approx 0.045 A}

Therefore, the current passing through the circuit is approximately 0.045 Amperes.

Step-by-Step Calculation

To provide a clear and concise understanding of the calculation process, let's break it down into steps:

  1. Calculate the total resistance of the resistors in series: Rresistors=55Ω+75Ω=130Ω{R_\text{resistors} = 55 \Omega + 75 \Omega = 130 \Omega}
  2. Calculate the total voltage of the cells in series: Vcells=4×1.5V=6V{V_\text{cells} = 4 \times 1.5 V = 6 V}
  3. Calculate the total internal resistance of the cells in series: Rinternal=4×0.8Ω=3.2Ω{R_\text{internal} = 4 \times 0.8 \Omega = 3.2 \Omega}
  4. Calculate the total resistance of the circuit: Rtotal=Rresistors+Rinternal=130Ω+3.2Ω=133.2Ω{R_\text{total} = R_\text{resistors} + R_\text{internal} = 130 \Omega + 3.2 \Omega = 133.2 \Omega}
  5. Apply Ohm's Law to calculate the current: I=VR=6V133.2Ω≈0.045A{I = \frac{V}{R} = \frac{6 V}{133.2 \Omega} \approx 0.045 A}

The Significance of Internal Resistance

As we've seen, the internal resistance of the cells plays a crucial role in determining the current flowing through the circuit. If we had ignored the internal resistance, we would have calculated the current as:

I=6V130Ω≈0.046A{I = \frac{6 V}{130 \Omega} \approx 0.046 A}

While the difference in this specific case is relatively small, it's important to understand that internal resistance can have a significant impact, especially in circuits with higher currents or lower external resistances. The internal resistance effectively reduces the voltage available to the external circuit, leading to a lower current flow.

In real-world applications, the internal resistance of batteries can also increase as they age or discharge, further affecting the performance of the circuit. This is why it's crucial to consider internal resistance when designing and analyzing electrical circuits, especially those powered by batteries or cells.

Conclusion

In conclusion, calculating the current in a series circuit involving resistors and cells requires a thorough understanding of series connections, Ohm's Law, and the concept of internal resistance. By carefully considering each component and applying the appropriate formulas, we can accurately determine the current flowing through the circuit. In this specific case, the current passing through the circuit with two resistors (55Ω and 75Ω) and four series-connected cells (1.5V each, 0.8Ω internal resistance) is approximately 0.045 Amperes. Understanding these principles is crucial for anyone working with electrical circuits, from students learning the basics to engineers designing complex systems.

This detailed explanation not only provides the answer to the specific problem but also delves into the underlying concepts and their significance, making it a valuable resource for understanding series circuits and Ohm's Law.