Question

Резистор сопротивлением 2 Ом подключен к источнику постоянного напряжения. Параллельно этому резистору подключили другой резистор сопротивлением 4 Ом. Как и во сколько раз изменилась мощность, выделяющаяся на участке цепи, состоящем из резисторов?

Answer 1

Если пренебречь внутренним сопротивлением источника напряжения
I1 = U/R
P1 = I1*U = U^2/R
I2 = U*(r+R)/(r*R)
P2 = I2*U = U^2*(r+R)/(r*R)
P2/P1 = {U^2*(r+R)/(r*R)} / {U^2/R} = (r+R)/r =  (4+2)/4 = 3/2
ответ, мощность увеличилась в 1,5 раза 

Answer 2

Calculation of Power in the Circuit

To determine how the power dissipated in the circuit has changed after connecting a second resistor in parallel, we need to consider the individual resistances and the formula for power in a circuit.

The power dissipated in a resistor can be calculated using the formula:

P = (V^2) / R

Where: - P is the power dissipated in watts (W) - V is the voltage across the resistor in volts (V) - R is the resistance of the resistor in ohms (Ω)

Let's calculate the power dissipated in the circuit before and after connecting the second resistor.

Power Dissipated Before Connecting the Second Resistor

Before connecting the second resistor, we have a single resistor with a resistance of 2 Ω connected to a constant voltage source. Let's assume the voltage across the resistor is V1.

The power dissipated in the single resistor can be calculated as:

P1 = (V1^2) / 2

Power Dissipated After Connecting the Second Resistor

After