Answer to Question #124912 in Classical Mechanics for Stefanie Hernandez-Mendez

Question

A battery has an emf of ε = 3 V, an internal resistance r = 29 Ω, and is connected to a resistor of R = 65 Ω.

(a) Express the current I through the circuit in terms of ε, r and R.
b. Calculate the numerical value of I in A.
c. Express the terminal voltage ΔV of the battery in terms of I and R.
d.Calculate the numerical value of ΔV in V.

Answer 1

Question (a)

Considering that the circuit is a series circuit, the equivalent resistance is

$R_{eq}=r+R$

Applying Ohm's Law you have to

$I=\frac{V}{R_{eq}}$

Where.

• The potential difference is $V=\varepsilon$
• The equivalent resistance is $R_{eq}=r+R$

The expression of the current is $\displaystyle \color{red}{\boxed{I=\frac{\varepsilon}{r+R}}}$

Question (b)

Numerically evaluating the voltage and resistance values

$I=\frac{3\;V}{29\;\Omega+65\;\Omega}=0.032\;A$

The numerical value of the current is $\displaystyle \color{red}{\boxed{I=0.032\;A}}$

Question (c)

Applying Ohm's Law, we have that the voltage at the terminals is

$\Delta V=I\;R$

Where.

• The current is $I$
• The connected resistance is $R$

The expression for the potential difference at the battery terminals is $\displaystyle \color{red}{\boxed{\Delta V=I\;R}}$

Question (d)

Numerical evaluation to calculate the potential difference

$\Delta V=I\;R$

$\Delta V=0.032\;A\times 65\;\Omega=2.08\;V$

The potential difference at the terminals is $\displaystyle \color{red}{\boxed{\Delta V=2.08\;V}}$