Easy V = IR problem, just help understand it.

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In summary, when a 9.0V battery is short-circuited, a 400mA current flows and the internal resistance of the battery can be calculated using the formula R_int = 9/(400*10^-3). This means that when a load is connected to the battery, the current flows through the load and also through the internal resistance of the battery. Short-circuiting a battery means connecting the poles with a wire with negligible resistance, which can cause the battery to become hot and drain quickly. The internal resistance of the battery includes the resistance of the electrodes and other materials within the battery.
  • #1
tnutty
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Homework Statement


When a 9.0V - battery is temporarily short-circuited, a 400mA- current flows.

What is the internal resistance of the battery?

Ans :R_int = 9/(400*10^-3)

Can someone explain why this is the answer, I understand its just V/I, but what exactly
does this mean? I am new to circuits so I am not sure what the term shot-circuit exactly
mean. I am also not sure what a internal resistance of the battery mean, just plain resistance?
 
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  • #2
To short-circuit a battery means that you connect the poles with a (short) piece of wire with negligible resistance. If you try it, you get even sparks, and your battery becomes hot, and flat in a very short time.
Now, internal resistance: When you connect something, for example a lamp to the battery, the current flows from one pole af the battery through the load (lamp) to the other pole and then through the battery to the first pole again, making a closed loop. But the battery itself has resistance. The electrodes themselves have, and there is some electrolyte material between them, and also some other material that absorbs gases, all of them have resistance. So we imagine a real battery as an ideal voltage source connected in series with a resistor Ri - the internal resistance of the battery. This internal resistance has to be taken into account if you calculate current. When you load the battery with a resistor R, R and Ri are connected in series, so I=E/(R+Ri). When the battery is short-circuited, R=0.

ehild
 
  • #3


I can provide a response to help you understand this problem. The formula V=IR represents Ohm's law, which relates voltage (V), current (I), and resistance (R) in a circuit. In this problem, we are given the voltage (9.0V) and the current (400mA) in a circuit with a battery. The internal resistance of a battery refers to the resistance within the battery itself, which can limit the flow of current.

When a battery is short-circuited, it means that the positive and negative terminals of the battery are connected directly, without any external resistance in the circuit. This allows for a large amount of current to flow through the battery, as there is no resistance to limit it.

The formula R_int = V/I is used to calculate the internal resistance of the battery. By substituting the given values into the formula, we can find the internal resistance of the battery. In this case, the answer is 22.5 ohms.

Understanding the concept of internal resistance is important in circuits, as it can affect the performance and lifespan of the battery. I hope this explanation helps you understand the problem better.
 

Related to Easy V = IR problem, just help understand it.

1. What is the Easy V = IR problem?

The Easy V = IR problem is a mathematical equation commonly used in electricity and electronics to calculate the relationship between voltage (V), current (I), and resistance (R). It states that the voltage across a circuit is equal to the current flowing through the circuit multiplied by the resistance of the circuit.

2. Why is the Easy V = IR problem important?

The Easy V = IR problem is important because it helps us understand how electricity flows through a circuit. By knowing the voltage, current, and resistance, we can calculate other important values such as power and energy. This equation is also essential for troubleshooting and designing circuits.

3. How do you solve the Easy V = IR problem?

To solve the Easy V = IR problem, you will need to know the values of two of the variables (voltage, current, or resistance) and use algebra to rearrange the equation to solve for the third variable. For example, if you know the voltage and resistance, you can use the formula I = V/R to calculate the current.

4. What are some real-life applications of the Easy V = IR problem?

The Easy V = IR problem is used in many real-life applications, such as designing electrical circuits, calculating power and energy usage in homes and buildings, and troubleshooting electronic devices. It is also essential in fields like engineering, physics, and electronics.

5. Are there any limitations to the Easy V = IR problem?

Yes, there are some limitations to the Easy V = IR problem. This equation assumes that the circuit is a simple series circuit and that the components have a linear relationship between voltage and current. In reality, circuits can be more complex, and some components may not follow a linear relationship, making this equation less accurate. Additionally, this equation does not take into account factors such as temperature and aging of components, which can affect the resistance and therefore the accuracy of the calculation.

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