Imagine we have two identical batteries with emf = 1.8V and an internal resistance of 0.6 ohms. Now we connect the batteries to a circuit with total resistance of 50 ohms. What is the maximum amount of power we can dissipate in the 50 ohm load? what is the minimum?
depends on how the batteries are connected, series or parallel.
Assuming series:
power=I^2(50)
where I= (2*1.8 )/51.2
To calculate the maximum and minimum power that can be dissipated in the 50-ohm load, we need to understand how the batteries behave in the circuit.
First, let's consider the batteries individually. Each battery has an EMF (electromotive force) of 1.8V and an internal resistance of 0.6 ohms. This means that when the battery is not connected to any circuit, it can provide a maximum potential difference of 1.8V across its terminals. However, when connected to a circuit, the internal resistance affects the actual potential difference delivered to the load.
In this case, since we have two identical batteries, we can connect them in either series or parallel. Let's examine both scenarios:
1. Series Connection:
When batteries are connected in series, their voltages add up, while the internal resistances also add up. Therefore, the series combination will have an EMF of 1.8V + 1.8V = 3.6V and an internal resistance of 0.6 ohms + 0.6 ohms = 1.2 ohms.
2. Parallel Connection:
When batteries are connected in parallel, their EMFs remain the same (1.8V), while the internal resistances become reciprocal. Therefore, the parallel combination will have an EMF of 1.8V and an internal resistance of 1/(1/0.6 ohms + 1/0.6 ohms) = 0.3 ohms.
Now, let's calculate the maximum and minimum power for each scenario:
1. Maximum Power:
To maximize power dissipation in the load, we need to ensure the load resistance matches the internal resistance of the batteries (i.e., the load resistance should be equal to the internal resistance of the series combination or the parallel combination, depending on the connection).
In the series connection, the load resistance that matches the internal resistance is 1.2 ohms. Therefore, the maximum power dissipated in the 50-ohm load can be calculated using the formula: P = (V^2) / R, where V is the potential difference and R is the resistance.
P_max (series) = ((3.6V)^2) / (50 ohms) ≈ 0.2592W
In the parallel connection, the load resistance that matches the internal resistance is 0.3 ohms. Therefore, the maximum power dissipated in the 50-ohm load can be calculated using the same formula:
P_max (parallel) = ((1.8V)^2) / (50 ohms) ≈ 0.065W
2. Minimum Power:
To minimize power dissipation in the load, we need to ensure the load resistance is significantly larger than the internal resistance of the batteries.
In both series and parallel connections, the load resistance of 50 ohms is already much larger than the internal resistance. Therefore, the minimum power dissipated in the 50-ohm load would occur in both configurations.
P_min = ((1.8V)^2) / (50 ohms) ≈ 0.065W
Thus, the maximum amount of power that can be dissipated in the 50-ohm load is approximately 0.2592W, and the minimum power is approximately 0.065W.