Question

Цепь состоит из источника тока с эдс 4,5 в и внутренним сопротивлением 1,5 Ом и проводников сопротивлением 4,5 Ом и 3 Ом соединенных последовательно.найти падение напряжения на R2

Answer 1

Ток в цепи составляет I=E/(rвн+R1+R2)=4.5/(1.5+4.5+3)=0.5 А. Теперь падение напряжения на резисторе 4.5 Ом будет Ur1=I*4.5=0.5*4.5=2.25 В. Падение на резисторе 3 Ом Ur2=I*3=0.5*3=1.5 В. В задаче не понятно, который резистор R1, а который R2 . Я взял первый как R1, а второй как R2

Answer 2

Calculation of Voltage Drop on R2

To find the voltage drop on resistor R2, we need to calculate the total resistance of the circuit and then use Ohm's Law to determine the voltage drop.

Given information: - The circuit consists of a current source with an electromotive force (EMF) of 4.5 V and an internal resistance of 1.5 Ω. - There are two resistors connected in series: R1 with a resistance of 4.5 Ω and R2 with a resistance of 3 Ω.

To calculate the total resistance of the circuit, we need to add the resistances of R1, R2, and the internal resistance of the current source:

Total resistance (R_total) = R1 + R2 + Internal resistance

Substituting the given values: R_total = 4.5 Ω + 3 Ω + 1.5 Ω R_total = 9 Ω

Now, we can use Ohm's Law to calculate the voltage drop on R2:

Voltage drop on R2 = Current * Resistance

The current flowing through the circuit can be calculated using Ohm's Law:

Current (I) = EMF / Total resistance

Substituting the given values: I = 4.5 V / 9 Ω I = 0.5 A

Now, we can calculate the voltage drop on R2:

Voltage drop on R2 = Current * Resistance of R2

Substituting the given values: Voltage drop on R2 = 0.5 A * 3 Ω Voltage drop on R2 = 1.5 V

Therefore, the voltage drop on resistor R2 is 1.5 V.