Question

A 9.0 V battery and three resistors are all connected in series. The potential difference across one resistor is measured to be 2.0 V, and the potential difference across another is measured to be 4.2 V. What is the potential difference across the third resistor

Answer 1

Answer:

The potential difference across the third resistor is $v_{3}=2.8\ V$.

Explanation:

We are told that the voltage of the battery is $v_{b}=9.0\ V$, that the potential difference across the first resistor is $v_{1}=2.0\ V$ and that the potential difference across the second resistor is $v_{2}=4.2\ V$.

If the three resistors are connected in series the sum of the potential difference over every resistor should be $v_{b}$ :

                                               $v_{b}=v_{1}+v_{2}+v_{3}$

                                       $9.0\ V=2.0\ V+4.2\ V+v_{3}$

                                       $v_{3}=9.0\ V-2.0\ V-4.2\ V$

                                                     $v_{3}=2.8\ V$

Answer 2

Answer:

2.8 V.

Explanation:

As shown in the diagram attached,

Since the resistors are connected in series, the total voltage from the battery is equal to sum of the  potential difference across each of the resistor.

V = v₁+v₂+v₃ ............... Equation 1

Where V = Emf of the battery, v₁ = potential difference across the first resistor, v₂ = potential difference across the second resistor, v₃ = potential difference across the third resistor

make v₃ the subject of the equation

v₃ = V-(v₁+v₂).............. Equation 2

Given: V = 9.0 V, v₁ = 2.0 V, v₂ = 4.2 V

Substitute into equation 2

v₃ = 9-(2+4.2)

v₃ = 9-6.2

v₃ = 2.8 V.