Question

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)

Answer 1

e)

Step-by-step explanation:

• Let's get first the values of the currents for the three cases.

i)

• 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:

`V = I*r_(i) + I*R_(1) (1)`

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

`I_(i) = (V)/(r_(int) + R_(i)) = (1.5V)/(0.1 \Omega + 1.4 \Omega) = 1.0 A (2)`

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:

`I_(ii) =(V_(term) )/(R_(ii) ) =(3.6V)/(1.8 \Omega) = 2.0 A (3)`

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:

`I_(iii) =(V-V_(term))/(r_(int) ) =(12.0 V- 11.0 V)/(0.2 \Omega) =(1.0V)/(0.2\Omega) = 5.0 A (4)`

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