Question

Резисторы сопротивления которых 2кОм и 3 кОм соединены последовательно и подключены к источнику постоянного напряжения 15В найдите силу тока в цепи и напряжение на каждом из резисторов

Answer 1

Закон ома для участка цепи в прикрепленном рисунке, т.к. соединение последовательное, то R=R1+R2=5*10^3 Ом
подставим в формулу закона Ома:
I=15/5*10^3=0,003=3*10^-3A
а теперь для каждого резистора подставим эту силу тока
U1=R1*I=2*10^3*3*10^-3=6B
U2=R2*I=3*10^3*3*10^-3=9B

Answer 2

Calculation of Current and Voltage in the Circuit

To find the current and voltage in the circuit, we need to apply Ohm's Law and the principles of series circuits.

Given: - Resistance of the first resistor (R1) = 2 kΩ (2000 Ω) - Resistance of the second resistor (R2) = 3 kΩ (3000 Ω) - Voltage of the power source (V) = 15 V

Step 1: Calculate the total resistance (R_total) in the circuit by summing the resistances of the individual resistors connected in series.

R_total = R1 + R2

Substituting the given values: R_total = 2000 Ω + 3000 Ω = 5000 Ω (5 kΩ)

Step 2: Calculate the current (I) flowing through the circuit using Ohm's Law.

I = V / R_total

Substituting the given values: I = 15 V / 5000 Ω = 0.003 A (3 mA)

Therefore, the current flowing through the circuit is 3 mA.

Step 3: Calculate the voltage drop across each resistor in the circuit.

In a series circuit, the voltage drop across each resistor is proportional to its resistance.

Voltage drop across R1 (V1) = I * R1

Substituting the given values: V1 = 0.003 A * 2000 Ω = 6 V

Voltage drop across R2 (V2) = I * R2

Substituting the given values: V2 = 0.003 A * 3000 Ω = 9 V

Therefore, the voltage across R1 is 6 V, and the voltage across R2 is 9 V.

Summary:

- The current flowing through the circuit is 3 mA. - The voltage across the 2 kΩ resistor (R1) is 6 V. - The voltage across the 3 kΩ resistor (R2) is 9 V.

Please note that the above calculations are based on the given information and assumptions of an ideal circuit.