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3 years ago

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51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

kΩ voltmeter across its terminals. (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio.

1 answer:

Tanya [424]3 years ago

5 0

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

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