How to Calculate Power Dissipated in a Parallel Battery Circuit with External Resistance

Understanding electrical circuits, particularly those involving batteries connected in parallel, requires a clear grasp of fundamental principles such as voltage, resistance, and current. In this article, we will delve into a specific scenario where 100 identical 6-volt batteries, each with an internal resistance of 1 ohm, are connected in parallel and a 6 ohm external resistor is connected to them. The goal is to determine the power dissipated by the external resistor. Let's break down the problem and solution in detail.

The Problem

The problem at hand is as follows: 100 identical batteries, each with a voltage of 6 volts and an internal resistance of 1 ohm, are connected in a parallel configuration. An external resistor of 6 ohms is also connected to this circuit. We need to find the power dissipated by this external resistor.

The Solution

Step 1: Calculate the Internal Resistance of the Parallel Connection

When batteries are connected in parallel, their internal resistances are divided. In this case, we have 100 batteries, each with an internal resistance of 1 ohm. Thus, the total internal resistance (( r_{total} )) can be calculated as:

[ r_{total} frac{1}{n} cdot R_{internal} ]

Where ( n ) is the number of batteries and ( R_{internal} ) is the internal resistance of each battery. Substituting the values:

[ r_{total} frac{1}{100} cdot 1 0.01 , text{ohms} ]

Step 2: Calculate the Net Resistance of the Circuit

The net resistance (( R_{net} )) of the circuit is the sum of the internal resistance of the batteries and the external resistance. Given that the internal resistance is now ( 0.01 , text{ohms} ) and the external resistance is ( 6 , text{ohms} ), we have:

[ R_{net} r_{total} R_{external} 0.01 6 6.01 , text{ohms} ]

Step 3: Calculate the Total Current Flowing Through the Circuit

Using the voltage of each battery (6 volts) and the net resistance calculated in the previous step, we can find the total current (( I )) using Ohm's Law:

[ I frac{V}{R_{net}} ]

[ I frac{6 , text{volts}}{6.01 , text{ohms}} approx 0.9983 , text{amperes} ]

Step 4: Calculate the Power Dissipated by the External Resistor

To find the power dissipated by the external resistor, we can use the formula for power (( P )) in an electrical circuit:

[ P V cdot I ]

Here, ( V ) is the voltage across the external resistor (which is 6 volts since the external resistor is connected directly to the 6-volt batteries), and ( I ) is the current flowing through the circuit. Thus:

[ P 6 , text{volts} cdot 0.9983 , text{amperes}^2 approx 6 cdot 0.9983 cdot 0.9983 approx 5.98 , text{watts} ]

Therefore, the power dissipated by the 6 ohm external resistor is approximately 5.98 watts.

Conclusion

The key to solving this problem lies in understanding the principle of parallel connections and how internal resistances are distributed. By breaking down the circuit and applying the relevant formulas, we can accurately determine the power dissipation in the external resistor.

Additional Insights

This problem highlights the importance of internal resistance in batteries and how it affects the overall performance of the circuit. Lowering the internal resistance is crucial for better output and efficiency. And understanding how to calculate power dissipation in such circuits is vital for various applications, from everyday electronics to more complex engineering projects.