Complete question:
In the circuit shown in the figure below (See image attached), suppose that the value of R1 is [tex] 500\,k\Omega [/tex]. To obtain a multimeter reading of 1 V between points B and C in the circuit, the value of R2 would have to be.
Answer:
[tex]R2=0.1\Omega [/tex]
Explanation:
First, we are going to the find current trough the circuit, because the resistors are on series the current is the same on each resistor so I=I1=I2. The Ohm's law for the circuit is:
[tex] V=R_{T}*I [/tex] (1) , with V the voltage of the battery (6V), I the current trough the circuit and [tex]R_{T} [/tex] the total resistance of the circuit, but for resistors on series the total resistance is the sum of the individual resistance so [tex] R_{T} = R1+R2[/tex] (2).
Using (2) on (1) and solving for I:
[tex]I=\frac{V}{R1+R2} [/tex] (3)
Ohm's law is true for the individual resistors too so we're going to apply that on R2:
[tex]V2=R2*I2 [/tex], but remember I2=I
[tex]V2=R2*I [/tex] (4), using (3) on (4)
[tex]V2=R2* \frac{V}{R1+R2} [/tex], solving for R2:
[tex] R1*V2+R2*V2=R2V[/tex]
[tex] R1*V2=R2(V-V2)[/tex]
[tex] R2=\frac{R1V2}{V-V2}=\frac{500\times10^{3}\Omega*1V}{6V-1V}[/tex], V2= 1V because we want that reading on the multimeter.
[tex]R2=100\,k\Omega [/tex]
