Question

to obtain a multimeter reading of 1 v between b and c in the circuit the value of r2 would have to be.

Answer 1

10k

Step-by-step explanation:

Answer 2

Complete question:

In the circuit shown in the figure below (See image attached), suppose that the value of R1 is
`500\,k\Omega`. 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:

`R2=0.1\Omega`

Step-by-step 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:

`V=R_(T)*I` (1) , with V the voltage of the battery (6V), I the current trough the circuit and
`R_(T)` the total resistance of the circuit, but for resistors on series the total resistance is the sum of the individual resistance so
`R_(T) = R1+R2` (2).

Using (2) on (1) and solving for I:

`I=(V)/(R1+R2)` (3)

Ohm's law is true for the individual resistors too so we're going to apply that on R2:

`V2=R2*I2`, but remember I2=I

`V2=R2*I` (4), using (3) on (4)

`V2=R2* (V)/(R1+R2)`, solving for R2:

`R1*V2+R2*V2=R2V`

`R1*V2=R2(V-V2)`

`R2=(R1V2)/(V-V2)=(500*10^(3)\Omega*1V)/(6V-1V)`, V2= 1V because we want that reading on the multimeter.

`R2=100\,k\Omega`