Question

10. A 1500 MS2 resistor is placed across the terminals of a battery, and a terminal voltage of 2.50 (1 point) V is measured. when this resistor is replaced with a 5.00 Ω resistor, the terminal voltage decreases to 1.75 V Find the terminal voltage when a 7.00 Ω resistor is used. O2.50 V 01.91 V 01.67 V O2.14 V

Answer 1

1.91 V is the correct answer to the given problem.

Let $V_i$ and r be the internal EMF and resistance of the battery.

For the 1500 $M\ \Omega$ resistance (say $R_1$), the equation for internal EMF of battery is,

$V_i = I_1R_1 + I_1r$

where $I_1$ is the current flowing in the circuit with $R_1$ resistance connected.

The current flowing in the circuit is given as,

$I_1 = \frac{V_1}{R_1}$

with $V_1$ as the terminal voltage in this case.

$I_1 = \frac{2.5}{1500 \times10^6}\\ = 1.67\times 10^{-9} A$

Similarly, for the 5 $\Omega$ resistance (say $R_2$), the equation for internal EMF of battery is,

$V_i = I_2R_2 + I_2r$

$I_2 = \frac{V_2}{R_2}$

$= \frac{1.75}{5}\\ = 0.35 A$

Equating both the equations to find r,

$I_1R_1 + I_1r = I_2R_2 + I_2r\\ r = \frac{I_1R_1- I_2R_2}{I_2 - I_1}$

$=\frac{2.5 - 1.75}{0.35 - 1.67 \times 10^{-9}}\\ = 2.14\ \Omega$

The internal EMF of the battery may be obtained from the second equation for $V_i$ as,

$V_i = 0.35*5 + 0.35*2.14\\ = 2.5\ V$

In the case of 7.00 $\Omega$ resistor (say R) connected across the battery, the current (I) in the circuit is given as,

$I = \frac{V_i}{R + r}\\ = \frac{2.5}{7.00+2.14}\\ = 0.27\ A$

The terminal voltage (V) in this case is,

$V = I*R\\ = 0.27*7\\ = 1.9\ V$