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GarryVolchara [31]

2 years ago

10

"If E = 7.50V and r=0.45Ω, find the minimum value of the voltmeter resistance RV for which the voltmeter reading is within 1.0%

of the emf of the battery. Express your answer numerically (in ohms) to at least three significant digits.

1 answer:

Virty [35]2 years ago

8 0

Answer:

The minimum value is $R_V =44.552\ \Omega$

Explanation:

From the question we are given that

The voltage is $E = 7.5V$

The internal resistance is $r = 0.45$

The objective of this solution is to obtain the minimum value of the voltmeter resistance for which the voltmeter reading is within 1.0% of the emf of the battery

What is means is that we need to obtain voltmeter resistance such that

V = (100% -1%) of E

Where E is the e.m.f of the battery and V is the voltmeter reading

i.e V = 99% of E = 0.99 E = 7.425

Generally

E = V + ir

where ir is the internal potential difference of the voltmeter and

V is the voltmeter reading

Making i the subject of the formula above

$i = \frac{(E-V)}{r}$

$=\frac{7.50-7.425}{0.45}$

$= 0.1667 A$

Now the current is constant through out the circuit so,

$V = iR_V$

Where $R_V$ is the value of voltmeter resistance

Hence $R_V = \frac{V}{i} = \frac{7.425}{0.1667}$

$=44.552\ \Omega$

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