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kaheart [24]
2 years ago
7

Discuss the magnitude of relative internal resistances of an ammeter and a voltmeter. Which meter is connected in series? in par

allel? Explain.
Physics
1 answer:
lorasvet [3.4K]2 years ago
8 0

The internal resistance of ammeter is zero and voltmeter is infinite and, ammeter is connected in series and the voltmeter is in parallel.

To find the answer, we have to study about the internal resistance.

<h3>What is the internal resistance of an ammeter and voltmeter?</h3>
  • A circuit has a parallel connection for the voltmeter and a series connection for the ammeter.
  • An ammeter is used to calculate the amount of current flowing through a circuit.
  • The value of the potential difference or voltage across the load is measured using a voltmeter (resistor).
  • The voltage reading on the voltmeter represents the amount of energy that each unit of charge has transmitted to the component.
  • An ideal ammeter should have zero internal resistance since it should permit current to flow through it.
  • To measure the current flowing through a circuit, an ammeter is connected in series with the circuit.
  • Since the internal resistance of the ideal voltmeter should prevent any current from passing through it, it is infinite.
  • Voltmeter measures the potential difference, it is connected in parallel.

Thus, we can conclude that, the internal resistance of ammeter is zero and voltmeter is infinite and, ammeter is connected in series and the voltmeter is in parallel.

Learn more about the internal resistance here:

brainly.com/question/24219708

#SPJ4

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Answer:

The length of the rod for the condition on the question to be met is L  =  1.5077 \ m

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is M  =  3.41 \ kg

           The mass of each small bodies is  m =  0.249 \ kg

           The moment of inertia of the three-body system with respect to the described axis is   I  =  0.929 \ kg \cdot  m^2

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        I =  I_r + 2 I_m

Where  I_r is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        I_r  =  \frac{ML^2 }{12}

And   I_m the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           I_m  =   m * [\frac{L} {2} ]^2 =  m*  \frac{L^2}{4}

Thus  2 *  I_m  =  2 *  m  \frac{L^2}{4}  = m  *  \frac{L^2}{2}

Hence

       I  =  M  *   \frac{L^2}{12}  +  m  * \frac{L^2}{2}

=>   I  =    [\frac{M}{12}  + \frac{m}{2}] L^2

substituting vales  we have  

        0.929   =    [\frac{3.41}{12}  + \frac{0.249}{2}] L^2

       L  =  \sqrt{\frac{0.929}{0.40867} }

      L  =  1.5077 \ m

     

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