Question

Calculate the terminal voltage of a 8.3100 V lithium cell in a digital wristwatch that draws 0.2800 mA and the cells internal resistance is 5.7500 1. Answer should be in the correct number of sig. figs. . V

Answer 1

Given,

the emf of the cell, $\varepsilon$ = 8.3100V

current drawn, I = 0.2800mA = (0.2800/1000) A = 0.0002800 A

internal resistance of the cell, r = 5.7500$\Omega$

Terminal voltage is the voltage across the terminals of a cell or battery.

When the cell is not connected in the circuit(when no current is flowing), the terminal voltage (V) of the cell, is equal its emf ($\varepsilon$).

When the cell is connected to a circuit(when current is flowing), the terminal voltage, V = $\varepsilon$ – Ir

V = 8.3100 – 0.0002800 x 5.7500

=  8.3100 – 0.001610

= 8.308390

V = 8.3084 V (up to proper significant figures)

So, the terminal voltage of the cell is 8.3084 V.