Conductors with a resistance of 4.8 and 8 ohms are connected in series and connected to a network

Conductors with a resistance of 4.8 and 8 ohms are connected in series and connected to a network with a voltage of 20V. determine the amperage.

R1 = 4.8 ohms.

R2 = 8 ohms.

U = 20 V.

I1 -?

I2 -?

I -?

With a series connection of conductors, the following formulas are valid: U = U1 + U2, I = I1 = I2, R = R1 + R2.

Find the total resistance of the entire circuit R: R = 4.8 ohms + 8 ohms = 12.8 ohms.

We express the current strength I in the entire circuit from Ohm’s law for a section of the circuit: I = U / R.

I = 20 V / 12.8 Ohm = 0.94 A.

With a series connection, the current strength in all sections of the circuit is the same: I = I1 = I2 = 0.94 A.

Answer: the current strength in all sections is the same and is I = I1 = I2 = 0.94 A.