Question

К неидеальному источнику с ЭДС E =12 В подключили параллельно соединённые между собой лампочку и идеальный вольтметр. При этом вольт- метр показал напряжение U =10 В, а на лампочке выделялась мощность P = 5 Вт . Найдите показание I идеального амперметра, включённого в данную цепь вместо вольтметра

Answer 1

P=U*I
значит ток в цепи через лампу и источник полампера - 5/10
на источнике падает 12-10 = два вольта
при найденном токе его сопротивление r = два вольта делить на полампера = 4 Ома

закоротив источник идеальным амперметром получим ток равным Е/r = 12/4= 3 ампера

Answer 2

Calculation of Current in the Circuit

To find the current I through the circuit, we can use Ohm's Law, which states that the current flowing through a conductor is equal to the voltage across the conductor divided by its resistance.

In this case, the circuit consists of a non-ideal voltage source with an electromotive force (EMF) of E = 12 V, a lamp, and an ideal voltmeter connected in parallel. The voltmeter measures a voltage of U = 10 V, and the lamp dissipates a power of P = 5 W.

We need to find the current I that would be measured by an ideal ammeter connected in place of the voltmeter.

To solve this problem, we need to make some assumptions and use the given information:

1. The voltage measured by the voltmeter is the voltage across the lamp and the voltmeter itself, as they are connected in parallel. 2. The power dissipated by the lamp is given by P = U * I, where U is the voltage across the lamp and I is the current flowing through it.

From the given information, we have:

- Voltage across the lamp and voltmeter: U = 10 V. - Power dissipated by the lamp: P = 5 W.

Using the power formula, we can rearrange it to solve for the current I:

P = U * I

Substituting the given values:

5 W = 10 V * I

Simplifying the equation:

I = 5 W / 10 V

Calculating the current:

I = 0.5 A

Therefore, the current measured by the ideal ammeter connected in place of the voltmeter would be 0.5 A.

Please note that this calculation assumes that the lamp and the voltmeter are the only components in the circuit and that the voltage source is non-ideal.