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3 years ago

9

Unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. part a a voltmeter with resistan

ce rv is connected across the terminals of a battery of emf e and internal resistance r. find the potential difference vmeter measured by the voltmeter.

2 answers:

Margaret [11]3 years ago

6 0

The EMF of the battery includes the force to to drive across its internal resistance. the total resistance:
R = internal resistance r + resistance connected rv
R = r + rv
Now find the current:
V 1= IR
I = R / V1
find the voltage at the battery terminal (which is net of internal resistance) using
V 2= IR
So the voltage at the terminal is:
V = V2 - V1
This is the potential difference vmeter measured by the voltmeter.

Bad White [126]3 years ago

6 0

Answer:

$\Delta V = \frac{E R_v}{R_v + r}$

Explanation:

A real voltmeter is connected to a non ideal battery of EMF "E"

here the resistance of the voltmeter is given as

$R = R_v$

also the internal resistance of the cell is given as

$R_{in} = r$

now the total resistance of the given combination of cell and the voltmeter is given as

$R = R_{in} + R_v$

$R = r + R_v$

now as per ohm's law the current is given as

$i = \frac{V}{R}$

$i = \frac{E}{R_v + r}$

now potential difference across the voltmeter is given as

$\Delta V = i R_v$

$\Delta V = \frac{E R_v}{R_v + r}$

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$\frac{T^2}{R^3}=k.......................(2)$

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Equation (2) therefore implies the following;

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We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

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Explanation:

Part 0

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