contestada

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.

Relax

Respuesta :

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance [tex]r= 5.00\ \Omega[/tex]

(a). We need to calculate the current

Using rule of loop

[tex]E-IR-Ir=0[/tex]

[tex]I=\dfrac{E}{R+r}[/tex]

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

[tex]I=\dfrac{3.200}{1.00\times10^{3}+5.00}[/tex]

[tex]I=3.184\times10^{-3}\ A[/tex]

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

[tex]V=E-Ir[/tex]

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

[tex]V=3.200-3.184\times10^{-3}\times5.00[/tex]

[tex]V=3.18\ V[/tex]

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

[tex]\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }[/tex]

[tex]\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}[/tex]

Hence, This is the required solution.