Analyzing A Series-Parallel Circuit Battery EMF 145V Internal Resistance 1 Ohm

This comprehensive guide delves into the analysis of a circuit featuring a battery with an electromotive force (EMF) of 145V and an internal resistance of 1 ohm, connected in series with a parallel combination of two resistors. We will explore the circuit diagram, calculate the combined resistance, determine the current flow, and analyze the voltage drop within the circuit.

1. Circuit Diagram: Visualizing the Electrical Network

To effectively analyze the circuit, let's begin by constructing a clear and concise circuit diagram. This visual representation will help us understand the interconnections of the various components and the flow of current. The circuit comprises a battery with an EMF (electromotive force) of 145V and an internal resistance of 1 ohm. This battery is connected in series with a parallel combination of two resistors. Let's denote the two resistors in parallel as R1 and R2. To completely define the circuit, we need to know the values of the resistors R1 and R2, so we will assume R1 = 4 ohms and R2 = 6 ohms for our calculations. In the circuit diagram, we represent the battery using the standard symbol, with the positive terminal clearly indicated. The internal resistance is shown as a resistor in series with the battery. The parallel combination of R1 and R2 is depicted with the resistors connected side-by-side, with their terminals joined together. The entire parallel combination is then connected in series with the battery. A well-drawn circuit diagram is the foundation for any circuit analysis. It provides a visual roadmap that guides us through the application of circuit laws and the determination of various electrical parameters. By accurately representing the components and their interconnections, we can avoid confusion and ensure that our calculations are based on a solid understanding of the circuit's structure.

2. Combined Resistance: Simplifying the Circuit

To determine the total current flowing in the circuit, we first need to calculate the combined resistance of the parallel combination of resistors R1 and R2. Resistors in parallel offer multiple paths for current to flow, effectively reducing the overall resistance. The formula for calculating the equivalent resistance (Rp) of two resistors in parallel is given by:

1/Rp = 1/R1 + 1/R2

Substituting the values of R1 = 4 ohms and R2 = 6 ohms, we get:

1/Rp = 1/4 + 1/6

1/Rp = (3 + 2) / 12

1/Rp = 5/12

Rp = 12/5 = 2.4 ohms

Therefore, the combined resistance of the parallel combination is 2.4 ohms. Now, we need to consider the internal resistance (r) of the battery, which is 1 ohm. Since the parallel combination is connected in series with the battery, the total resistance (Rtotal) of the circuit is the sum of the parallel resistance (Rp) and the internal resistance (r):

Rtotal = Rp + r

Rtotal = 2.4 ohms + 1 ohm

Rtotal = 3.4 ohms

Thus, the total resistance of the circuit is 3.4 ohms. This value represents the effective opposition to current flow offered by the entire circuit. Knowing the total resistance is crucial for calculating the current and voltage drops within the circuit.

3. Current in the Circuit: Applying Ohm's Law

Now that we have determined the total resistance of the circuit, we can calculate the current flowing through it using Ohm's Law. Ohm's Law states that the current (I) flowing through a conductor is directly proportional to the voltage (V) across it and inversely proportional to the resistance (R). Mathematically, this is expressed as:

I = V / R

In our circuit, the voltage is the EMF of the battery, which is 145V, and the resistance is the total resistance of the circuit, which we calculated as 3.4 ohms. Substituting these values into Ohm's Law, we get:

I = 145V / 3.4 ohms

I ≈ 42.65 Amperes

Therefore, the current flowing in the circuit is approximately 42.65 Amperes. This current represents the rate at which charge is flowing through the circuit. It is a crucial parameter for understanding the circuit's behavior and for determining the power dissipated by the various components. The current will be the same throughout the series portion of the circuit, including the battery and the combined parallel resistance.

4. Lost Volt: Internal Voltage Drop

The lost volt, also known as the internal voltage drop, is the voltage drop across the internal resistance of the battery. This voltage drop occurs because the battery itself has an internal resistance that opposes the flow of current. The lost volt reduces the actual voltage available to the external circuit. To calculate the lost volt (Vlost), we can use Ohm's Law again:

Vlost = I * r

Where I is the current flowing through the circuit (42.65 Amperes) and r is the internal resistance of the battery (1 ohm). Substituting these values, we get:

Vlost = 42.65 Amperes * 1 ohm

Vlost ≈ 42.65 Volts

Therefore, the lost volt in the circuit is approximately 42.65 Volts. This means that out of the battery's EMF of 145V, 42.65V is dropped across the internal resistance, leaving 102.35V (145V - 42.65V) available to the external circuit, specifically the parallel combination of resistors R1 and R2. The lost volt is an important consideration in circuit design, as it affects the overall efficiency of the circuit and the voltage available to power external loads. A higher internal resistance leads to a larger lost volt and a lower voltage available to the external circuit.

In summary, we have analyzed a circuit comprising a battery with an EMF of 145V and an internal resistance of 1 ohm, connected in series with a parallel combination of two resistors (R1 = 4 ohms and R2 = 6 ohms). We have determined the combined resistance of the parallel combination (2.4 ohms), the total resistance of the circuit (3.4 ohms), the current flowing in the circuit (42.65 Amperes), and the lost volt (42.65 Volts). This analysis provides a comprehensive understanding of the circuit's behavior and the interplay of its various components.

Keywords Review and Questions Clarification

Let's address the keywords and ensure the questions are clear and well-defined:

• Original Keywords: A battery of emf 145 and literal resistance $1 \sim$ is commented in series with aparallel connection if 2 resistors.

1. Draw a circuit diagram
2. Calculate combined resistors
3. Calculate current in the circuit
4. Calculate the lost volt in the

• Repair Input Keyword: Analyzing a series-parallel circuit with a battery of EMF 145V and internal resistance 1 ohm connected to a parallel combination of two resistors. Find: 1) Draw the circuit diagram. 2) Calculate the combined resistance of the resistors. 3) Calculate the current in the circuit. 4) Calculate the lost voltage (internal voltage drop).

The revised keywords and questions provide a more precise and easily understandable description of the problem.