Question

In the circuit shown in Figure 4 below, how much resistance should be connected to a 10 Ω resistor so that the points M and N are at the same potential?

Answer 1

For points M equal to the potential at point N (V_M = V_N), the circuit must satisfy the condition of a balanced Wheatstone bridge.

In a balanced bridge, the ratio of the resistances in the top arms must equal the ratio of the resistances in the bottom arms.

Let R₁, R₂, R₃ and R₄ be the resistors in the bridge:

R₁ (Top  left) = 3 Ω

R₂ (Top right) = 6 Ω

R₃ (Bottom left) = 4 Ω

R₄ (Bottom right) = The target equivalent resistance for the 10 Ω branch.

`(R_1)/(R_2) = (R_3)/(R_4)`

`(3 Ω)/(6 Ω) = (4 Ω)/(R_4)`

`1/2 = (4 Ω)/(R_4)`

R₄ = 8 Ω

To obtain a total resistance of 8 Ω from an existing 10 Ω resistor, another resistor (R_x​) must be connected in parallel, since the required resistance is less than the given resistance.

Using the parallel resistance formula:

`1/(R_x) = 1/10 + 1/(R_p)`

`1/8= 1/10 + 1/(R_p)`

`1/(R_p) = 1/8 - 1/10`

= `(10 - 8)/80`

= `2/80`

= `1/40`

∴ R_p = 40 Ω