The circuit section consists of four resistors. Resistors R1 = 2 ohms and R2 = 3 ohms are connected in parallel.

| March 9, 2021 | education

The circuit section consists of four resistors. Resistors R1 = 2 ohms and R2 = 3 ohms are connected in parallel. Resistors R3 = 3 Ohm and R4 = 0.8 Ohm are connected in series with them. A voltage of U = 20 V is applied to the ends of the section. Find the current in each of the four resistors.

To solve this problem, it is necessary to apply Ohm’s law.

I = U / R

When connecting conductors in series:

Utot. = U1 + U2

Itot. = I1 = I2

Rtot. = R1 + R2

When connecting conductors in parallel:

Utot. = U1 = U2

Itot. = I1 + I2

Rtot. = (R1 * R2) / (R1 + R2)

Let us bring to the equivalent resistances:

Req1 = (R1 * R2) / (R1 + R2) = (2 * 3) / (2 + 3) = 1.2 Ohm

Req2 = R3 + R4 = 3 + 0.8 = 3.8 Ohm

Req3 = Req1 + Req2 = 1.2 + 3.8 = 5 Ohm

I = U / R = 20/5 = 4 A

Accordingly, on R3 and R4, the current is I = 4 A, i.e. I3 = I4 = 4 A

Checking.

Utot = U1 + U2 = I * Req1 + I * Req2 = 4 * 1.2 + 4 * 3.8 = 20 V

Further.

U1 = I * Req1 = 4 * 1.2 = 4.8 V

I1 = U1 / R1 = 4.8 / 2 = 2.4 A

I2 = U1 / R2 = 4.8 / 3 = 1.6 A

Checking the balance of currents:

I1 + I2 = I3 = 4 —> 2.4 + 1.6 = 4 = 4 —> 4 = 4 = 4

Voltage balance check:

4.8 + 4 * 3.8 = 20 —> 20 = 20

Answer: I1 = 2.4 A; I2 = 1.6 A; I3 = I4 = 4 A



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