A circuit question involving internal resistances

In summary, the conversation is about a battery with an emf of 6v connected in series with an ammeter and a voltmeter. When a resistor is connected in parallel with the voltmeter, the voltmeter reading decreases by a factor of 3 and the ammeter reading increases by the same factor. The question is asking for the initial reading of the voltmeter, with all elements except the battery having unknown internal resistances. The conversation includes discussions about using basic rules about resistors, Kirchhoff's rules, and eventually simplifying equations to get the correct answer of 4.5v for the initial voltmeter reading.
  • #1
jianghan
25
0

Homework Statement



A battery has an emf of 6v. The battery is connected in series with an ammeter and a voltmeter.If a certain resistor is connected in parallel with the voltmeter,the voltmeter reading decreases by a factor of 3,and the ammeter reading increases by the sme factor. What is the initial reading V of the voltmeter? All elements except battery in the circuit have unknown internal resistances

2. The attempt at a solution
R3 is the resistor that is connected in parallel with the voltmeter.
I also use R1,R2 to represent internal resistances of voltmeter and ammeter which i assume are resistors connected in series to the voltmeter and ammeter.(ammeter then measures current flowing through R2 while voltmeter measures pd across R1 or R3)

I tried to construct equations using basic rules about resistors connected in series or parallel,after solving some (overly complex) equations,i had V=4.8v while the answer is v=4.5v

I also tried to use Kirchhoff's rules to solve,but it did not work out.
 
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  • #2
Can you post your equations? I was able to get 4.5V using voltage divider, KCL, and ohms law (x2).

I really just started writing equations until something popped out at me - then I had an idea where to go!
 
  • #3
tindel said:
Can you post your equations? I was able to get 4.5V using voltage divider, KCL, and ohms law (x2).

I really just started writing equations until something popped out at me - then I had an idea where to go!

I simplified my equations and now i can get the correct answer of 4.5v
my equations are as follows
assume R1//R3 is r

IR1+IR2=6V
3IR2+3Ir=6V
solve these to get R1-2R2=3r

one more equation:
IR1=3(3I)r → R1=9r

so R1=3R1-6R2
R1=3R2

therefore IR1=4.5V
 

Related to A circuit question involving internal resistances

1. What is meant by internal resistance in a circuit?

Internal resistance refers to the resistance within a power source, such as a battery or generator, that limits the flow of current in a circuit. It is caused by the resistance of the materials used to make the power source and can affect the overall performance of the circuit.

2. How is internal resistance calculated?

The internal resistance of a power source can be calculated by measuring the voltage drop across the source and the current flowing through it, and using the formula R = V/I, where R is the internal resistance, V is the voltage, and I is the current.

3. What is the effect of internal resistance on a circuit?

Internal resistance can cause a decrease in the voltage and power output of a circuit, as some of the energy is lost within the power source. It can also affect the stability and efficiency of the circuit, as well as the accuracy of voltage measurements.

4. How does internal resistance affect the behavior of a battery in a circuit?

The internal resistance of a battery can cause the voltage to drop significantly when a high current is drawn from it. This can lead to a decrease in the battery's capacity and a shorter lifespan. However, in some cases, a higher internal resistance can also provide protection against short circuits and overloading.

5. Can internal resistance be reduced or eliminated in a circuit?

Internal resistance cannot be completely eliminated, as it is inherent in the materials used to make the power source. However, it can be reduced by using materials with lower resistance and by optimizing the design of the circuit. Some circuits also use external resistors to compensate for the effects of internal resistance.

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