Question

в цепь включены последовательно три проводника с сопротивлениями 5,6,12 Ом. Какую силу тока показывает амперметр,а каково напряжение на третьем проводнике ,если падение напряжения на всем участке равно 4,6В?

Answer 1

Calculation of Current and Voltage in the Circuit

To calculate the current and voltage in the given circuit, we need to consider the resistances of the three conductors and the total voltage drop across the circuit.

The circuit consists of three conductors connected in series, each with resistances of 5 Ω, 6 Ω, and 12 Ω respectively. Let's denote these resistances as R1, R2, and R3.

The total resistance in a series circuit is the sum of the individual resistances. Therefore, the total resistance in this circuit is:

R_total = R1 + R2 + R3

Substituting the given values, we have:

R_total = 5 Ω + 6 Ω + 12 Ω = 23 Ω

Now, let's denote the total voltage drop across the circuit as V_total, which is given as 4.6 V.

According to Ohm's Law, the current flowing through a circuit is equal to the voltage divided by the resistance:

I = V / R

Using this formula, we can calculate the current flowing through the circuit:

I = V_total / R_total

Substituting the given values, we have:

I = 4.6 V / 23 Ω = 0.2 A

Therefore, the ammeter will show a current of 0.2 A in the circuit.

To calculate the voltage across the third conductor, we need to consider the voltage drop across each resistor. Since the resistors are connected in series, the voltage drop across each resistor is proportional to its resistance.

Let's denote the voltage drop across the first, second, and third resistors as V1, V2, and V3 respectively.

Since the total voltage drop across the circuit is 4.6 V, we can write the following equation:

V_total = V1 + V2 + V3

We can also express the voltage drop across each resistor using Ohm's Law:

V1 = I * R1 V2 = I * R2 V3 = I * R3

Substituting these equations into the previous equation, we have:

V_total = I * R1 + I * R2 + I * R3

Factoring out the common factor of I, we get:

V_total = I * (R1 + R2 + R3)

Substituting the given values, we have:

V_total = 0.2 A * (5 Ω + 6 Ω + 12 Ω) = 0.2 A * 23 Ω = 4.6 V

Therefore, the voltage across the third conductor is 4.6 V.

Please let me know if you need any further clarification or assistance!