Four electric resistors $$ \mathrm{A}, \mathrm{B}, \mathrm{C} $$, and D , having resistances of $$ 2 \Omega, 4 \Omega, 6 \Omega $$ and respectively, are connected in parallel with a 12 V battery. If 1.5 A of electric curr is flowing in resistor B , find out the average resistance of the circuit and the amo of voltage in resistor B . Determine the average current flowing through the circui well.

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We will show that if the four resistors

A = 2 Ω, B = 4 Ω, C = 6 Ω, D = 12 Ω

are “arranged” with a 12‐V supply so that a measured 1.5 A flows in resistor B, then the circuit “acts” as a series connection of two equal resistances (one being the parallel combination of A, B, C, and D and the other an “internal” or extra resistance) so that the total (or “average”) resistance is 2 Ω, the drop on resistor B is 6 V, and the total (or “average”) current is 6 A.

Below is one way to arrive at that conclusion.

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A. A Possible Interpretation of the Problem

At first glance the statement “Four electric resistors A, B, C and D … are connected in parallel with a 12 V battery” would seem to imply that each resistor is directly across 12 V. But then by Ohm’s law resistor B (4 Ω) would carry
  I = V/R = 12/4 = 3 A.
Since we are told that only 1.5 A flows in resistor B the only way to “explain” this is that the resistors are not connected directly across the 12 V but rather form a sub‐circuit that is fed from the battery through an extra resistance. (Sometimes textbooks call the overall resistance “the average resistance of the circuit” meaning the effective or equivalent resistance seen by the battery.) One standard arrangement is as follows:

12 V battery → R_series → (parallel combination of A, B, C, and D)

In such a two–stage voltage divider the voltage “fed” to the parallel part is lower than 12 V.

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B. Determining the Voltage on Resistor B

In a parallel branch every resistor sees the same voltage. Let this voltage be Vp. We are told that in branch B the current is

I_B = Vp/4      [Ohm’s law]

and we are told I_B = 1.5 A. Hence

Vp = (1.5 A)(4 Ω) = 6 V.

That is, the voltage across resistor B (and hence across the entire parallel network) is 6 V.

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C. Deducing the Extra (Series) Resistor

Since the battery is 12 V and the parallel network only “sees” 6 V the remaining 6 V must drop across a resistor in series with the parallel group. Let this series resistor be R_s. Then the total voltage is divided as

12 V = V_s + V_p   with V_p = 6 V → V_s = 6 V.

The current in the whole circuit is the same through the series resistor and is given by

I_total = 6 V / R_s.

On the other hand, the effective or equivalent resistance of the parallel portion is

R_p = 1/(1/2 + 1/4 + 1/6 + 1/D)
    = 1/(½ + ¼ + 1/6 + 1/D).

Now, for the numbers to work (that is, for resistor B to have 6 V across it giving 1.5 A and for the total battery voltage to be 12 V) the extra resistor R_s must be equal to the effective parallel resistance R_p. In many textbook problems an additional “drop resistor” is chosen so that

R_s = R_p.

Then the voltage divider gives

V_p = [R_p/(R_s + R_p)] (12 V) = [R_p/(2R_p)] (12 V) = 6 V,
which is exactly what we have.

So we must have R_s = R_p.

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D. Finding the Value of D

Often such a problem is “designed” so that the numbers come out neatly. In our case, let us choose D such that

1/(2 Ω) + 1/(4 Ω) + 1/(6 Ω) + 1/(D) = 1/R_p,
with R_p coming out to be a simple number.

A very common choice is to take D = 12 Ω. Let’s check:

1/2 = 0.5
  1/4 = 0.25
  1/6 ≈ 0.1667
  1/12 ≈ 0.0833

Adding these yields

0.5 + 0.25 + 0.1667 + 0.0833 = 1.0000.

Thus

R_p = 1/(1) = 1 Ω.

Since we require R_s = R_p then R_s = 1 Ω.
The total (or “average”) resistance of the entire circuit seen by the battery is then

R_total = R_s + R_p = 1 Ω + 1 Ω = 2 Ω.

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E. The Overall Circuit Current

With an equivalent total resistance of 2 Ω and a battery of 12 V the overall current from the supply is

I_total = 12 V / 2 Ω = 6 A.
This is the "average current flowing through the circuit."

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F. Summary of Answers

With D chosen as 12 Ω, the circuit (with a series resistor R_s = 1 Ω feeding a parallel network of the four resistors) works out as follows:

1. The effective (or
The average resistance of the circuit is 2 Ω, the voltage across resistor B is 6 V, and the average current flowing through the circuit is 6 A.

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