On a cold day a battery has a terminal voltage of Vi- 11.96 V and an internal resistance of ri = 0.0095 ? On a warm day the battery has a terminal voltage of V2-1161 V and an internal resistance of r2 -0.021 2. The device it powers draws the same current at both tempertures. Assume the emf of the battery does not depend on temperature. 33% Part (a) Enter an expression for the current, I, in terms of the given quantities.Part (b) Calculate the cument, I, in amperes. 33% Part (c) what is the battery’s emf, in volts?
Answer:
(a) [tex]I = \dfrac{V_1-V_2}{r_2-r_1}[/tex]
(b) 29.91 A
(c) 12.24 V
Explanation:
(a) The emf of a battery is the sum of its terminal voltage and lost voltage. The lost voltage is due to the internal resistance. By this, we have for the first condition,
[tex]E = V_1 + Ir_1[/tex]
E is the emf, [tex]V_1[/tex] is the terminal voltage and [tex]Ir_1[/tex] is the lost voltage with I being the current.
Since the same current is drawn in both conditions and the emf does not depend on temperature, we have similar for the second condition
[tex]E = V_2 + Ir_2[/tex]
Equating both equations and solving for I, we have
[tex]E = V_1 + Ir_1 = V_2 + Ir_2[/tex]
[tex]I = \dfrac{V_1-V_2}{r_2-r_1}[/tex]
(b) Substitute the values from the question into the expression for I in (a)
[tex]I = \dfrac{11.96-11.61}{0.0212-0.0095}=29.91\text{ A}[/tex]
(c) Substitute the value of I into any of the equations of E
[tex]E = V_1 + Ir_1[/tex]
[tex]E = 11.96 + 29.91\times0.0095 = 12.24\text{ V}[/tex]