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Lemur [1.5K]
3 years ago
7

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.
Physics
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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