Rank the following circuits in order from highest to lowest values of the current in the circuit.

i. a 1.4-Ω resistor connected to a 1.5-V battery that has an internal resistance of 0.10 Ω;
ii. a 1.8-Ω resistor connected to a 4.0-V battery that has a terminal voltage of 3.6 V but an unknown internal resistance;
iii. an unknown resistor connected to a 12.0-V battery that has an internal resistance of 0.20 Ω and a terminal voltage of 11.0 V.

a. (i), (iii), (ii)
b. (iii), (i), (ii)
c. (ii), (iii), (i)
d. (i), (ii), (iii)
e. (iii), (ii), (i)
f. (ii), (i), (iii)

The battery forms a series circuit with its internal resistance and the 1.4 Ω resistor. Since the current is the same at any point of the circuit, and the sum of all voltages along a closed circuit must be zero, we can apply Ohm's Law in each resistor, as follows:

[tex]V = I*r_{i} + I*R_{1} (1)[/tex]

Replacing V, ri and R₁ by their values, we can solve for the current I as follows:

[tex]I_{i} = \frac{V}{r_{int} + R_{i}} = \frac{1.5V}{0.1 \Omega + 1.4 \Omega} = 1.0 A (2)[/tex]

ii)

Since the voltage of the battery is 4.0 V (open circuit voltage), and it falls to 3.6 V when is connected to a 1.8Ω resistor, this means that the voltage through the resistor must be 3.6 V, due to the sum of all voltages along a closed circuit must be zero.
So, we can find the current through the circuit, applying Ohm's Law to the 1.8Ω resistor, as follows:

[tex]I_{ii} =\frac{V_{term} }{R_{ii} } =\frac{3.6V}{1.8 \Omega} = 2.0 A (3)[/tex]

iii)

Since the 12.0 V battery has a terminal voltage of 11.0 , this means that the voltage through the internal resistance of 0.2 Ω, must be 1.0 V.
So we can find the current Iiii, applying Ohm's Law to the internal resistance value, as follows:

[tex]I_{iii} =\frac{V-V_{term}}{r_{int} } =\frac{12.0 V- 11.0 V}{0.2 \Omega} =\frac{1.0V}{0.2\Omega} = 5.0 A (4)[/tex]

So, the highest current is the Iiii, followed by Iii and Ii, which is stated by e).