Question

ЭДС источника точка 10 В, внутреннее сопротивление 20 м, сопротивление нагрузки R=18Ом. Найти силу тока в цепи, напряжение на нагрузке, мощность тока на нагрузке.

Answer 1

I=E/(R+r)=10/(18+2)=0,5 A
U=I*R=0,5*18=9 B
P=I^2*R=0,5^2*18=4,5 Вт

Answer 2

Problem Statement

We are given an electromotive force (EMF) source with a voltage of 10 V, an internal resistance of 20 mΩ, and a load resistance of 18 Ω. We need to find the current flowing through the circuit, the voltage across the load, and the power dissipated by the load.

Solution

To solve this problem, we can use Ohm's Law and the concept of equivalent resistance.

Step 1: Equivalent Resistance The total resistance in the circuit is the sum of the internal resistance and the load resistance. Therefore, the equivalent resistance (R_eq) is given by:

R_eq = R_load + R_internal

Substituting the given values:

R_eq = 18 Ω + 20 mΩ

R_eq = 18.02 Ω

Step 2: Current Using Ohm's Law, we can find the current (I) flowing through the circuit:

I = V / R_eq

Substituting the given voltage:

I = 10 V / 18.02 Ω

I ≈ 0.554 A

Therefore, the current flowing through the circuit is approximately 0.554 A.

Step 3: Voltage across the Load The voltage across the load (V_load) can be found using Ohm's Law:

V_load = I * R_load

Substituting the given values:

V_load = 0.554 A * 18 Ω

V_load ≈ 9.972 V

Therefore, the voltage across the load is approximately 9.972 V.

Step 4: Power Dissipated by the Load The power dissipated by the load (P_load) can be calculated using the formula:

P_load = I^2 * R_load

Substituting the given values:

P_load = (0.554 A)^2 * 18 Ω

P_load ≈ 5.84 W

Therefore, the power dissipated by the load is approximately 5.84 W.

Summary

In summary, for the given circuit with an EMF source of 10 V, an internal resistance of 20 mΩ, and a load resistance of 18 Ω, we have determined the following: - The current flowing through the circuit is approximately 0.554 A. - The voltage across the load is approximately 9.972 V. - The power dissipated by the load is approximately 5.84 W.