Internal resistance and electromotive force problem

In summary, the conversation discusses a complete circuit with a battery, resistor, and switch, and the readings of a voltmeter and ammeter when the switch is open and closed. It also mentions the concept of internal resistance and using Ohm's law to calculate voltage.
  • #1
nutzweb
12
0
help me with this please... i just couldn't get it.

1) a complete circuit consists of a 12.0 V battery with a 4.50 ohm resistor and switch. the internal resistance of the battery is 0.30 ohms when switch is open. what does the voltmeter read when placed:

a. across the terminal of the battery when the switch is open
b. across the resistor when the switch is open
c. across the terminal of the battery when the switch is closed
d. across the resistor when the switch is closed

2) when the switch S is open, the voltmeter V reads 2.0 V. when the switch is closed, the voltmeter reading drops to 1.50 V and the ammeter reads 1.20 A. find the emf (electromotive force) and the internal resistance of the battery. assume that the two meters are ideal so they don't affect the circuit.
 
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  • #2
nutzweb said:
help me with this please... i just couldn't get it.

1) a complete circuit consists of a 12.0 V battery with a 4.50 ohm resistor and switch. the internal resistance of the battery is 0.30 ohms when switch is open. what does the voltmeter read when placed:

a. across the terminal of the battery when the switch is open
b. across the resistor when the switch is open
c. across the terminal of the battery when the switch is closed
d. across the resistor when the switch is closed

2) when the switch S is open, the voltmeter V reads 2.0 V. when the switch is closed, the voltmeter reading drops to 1.50 V and the ammeter reads 1.20 A. find the emf (electromotive force) and the internal resistance of the battery. assume that the two meters are ideal so they don't affect the circuit.
V=IR. Ohm's law.

AM
 
  • #3


First, let's define internal resistance and electromotive force (emf). Internal resistance is the resistance within the battery itself, which can affect the voltage output of the battery. Electromotive force is the voltage produced by the battery, which is equal to the potential difference between the positive and negative terminals.

Now, let's tackle the first question. When the switch is open, it means that there is no current flowing in the circuit. This means that the voltmeter will read the full emf of the battery, which in this case is 12.0 V. This is because there is no voltage drop across the internal resistance since there is no current flowing through it. Therefore, the voltmeter will read 12.0 V when placed across the terminals of the battery.

When the switch is open, the voltmeter will read the voltage drop across the resistor, which is calculated using Ohm's Law (V = IR). Plugging in the values given, the voltmeter will read 1.2 V (0.3 ohms x 4.5 ohms = 1.2 V).

Now, let's move on to the second question. We are given the voltmeter reading and ammeter reading when the switch is open and closed. When the switch is open, the voltmeter reads the full emf of the battery, which we can calculate by rearranging Ohm's Law (V = IR) to solve for emf. Plugging in the values given, we get an emf of 2.0 V.

When the switch is closed, the voltmeter reading drops to 1.5 V. This is because there is now a voltage drop across both the internal resistance and the resistor. Using Kirchhoff's Voltage Law (KVL), we can set up an equation: emf - IR - IR = 1.5 V. Plugging in the values and solving for emf, we get an emf of 2.5 V.

Using the two values of emf (2.0 V and 2.5 V), we can solve for the internal resistance of the battery. We can set up another equation using Kirchhoff's Voltage Law: emf - IR = V. Plugging in the values and solving for R, we get an internal resistance of 0.5 ohms.

In conclusion, internal resistance and electromotive force can affect the voltage output
 

1. What is internal resistance and electromotive force (EMF)?

Internal resistance is the resistance that exists within a power source, such as a battery, that limits the flow of current. Electromotive force (EMF) is the potential difference between two points in a circuit, which is responsible for creating the flow of current.

2. How do internal resistance and EMF affect the performance of a circuit?

Internal resistance can cause a decrease in the overall voltage and current of a circuit, leading to a decrease in the performance of the circuit. EMF, on the other hand, is necessary for the circuit to function and is responsible for maintaining a constant flow of current.

3. How do you calculate internal resistance and EMF?

Internal resistance can be calculated by measuring the voltage drop across a resistor in the circuit and using Ohm's law (R = V/I) to determine the resistance. EMF can be calculated by measuring the potential difference between two points in the circuit.

4. What factors can affect the internal resistance and EMF of a power source?

The internal resistance of a power source can be affected by the type of material used in the construction of the source, its size and shape, and the temperature. EMF can be affected by the type of chemical reaction happening within the power source, the materials used, and the number of cells connected in the circuit.

5. How can you minimize the effects of internal resistance and EMF in a circuit?

The effects of internal resistance can be minimized by using a power source with a lower resistance or by using a larger conductor to reduce the resistance. EMF can be maintained by using a power source with a higher EMF, using a larger number of cells in the circuit, or using a larger conductor to reduce the resistance.

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