Three conductors with a resistance equal to 10.20.30 ohms are connected in series.

Three conductors with a resistance equal to 10.20.30 ohms are connected in series. The voltage on 1 conductor is 10 V. Determine the voltage on each conductor

First, we find the current that flows through these conductors. Since the conductors are connected in series, the current passing through each of them will be the same. I.e:

I = I1 = I2 = I3,

where I1 is the current in the first conductor, I2 is the current in the second conductor, I3 is the current in the second conductor, I is the total current.

Since the conductors are connected in series, the total resistance will be determined by the formula:

R = R1 + R2 + R3,

respectively

R = R1 + R2 + R3 = 10 + 20 + 30 = 60 (Ohm)

Following Ohm’s law, we determine the total current in the circuit:

I = U / R;

I = U / R = 10/60 = 1/6 ≈ 0.167 (A).

Let’s determine the voltages on each conductor using Ohm’s law:

U = I * R,

respectively

U1 = I * R1, U2 = I * R2, U3 = I * R3;

U1 = I * R1 = 0.167 * 10 = 1.67 (B);

U2 = I * R2 = 0.167 * 20 = 3.34 (B);

U3 = I * R3 = 0.167 * 30 = 5.01 (B);

Answer: U1 = 1.67 (V); U2 = 3.34 (B); U3 = 5.01 (B);