Question

Две лампочки сопротивлениями 12 Ом и 8 Ом соединены последовательно и подключены к источнику с напряжением 4 В. Определите силу тока в цепи и напряжение на каждой лампочке.

Answer 1

ДАНО: R₁=12 Ом,  R₂=8 Ом, U=4В.
При последовательном соединении общее сопротивление R=R₁+R₂,
R= 12+8 =20 Ом.
Так как напряжение U=4В, можно найти силу тока по закону Ома I=U/R,
I=4В/20 Ом=0,2 А . При последовательном соединении I=I₁=I₂, U=U₁+U₂,
U₁=I₁*R₁, U₁= 0,2*12=2,4 В и U₂=I₂*R₂, U₂=0,2*8=1,6 В (или для
 проверки :  4В-2,4 В=1,6 В ).
Ответ: 0,2 А,  2,4 В, 1,6 В.

Answer 2

Calculation of Current and Voltage in the Circuit

To determine the current and voltage in the circuit, we need to apply Ohm's law and the rules for series circuits.

Given: - Resistance of the first lamp (R1) = 12 Ω - Resistance of the second lamp (R2) = 8 Ω - Voltage of the source (V) = 4 V

Step 1: Calculate the total resistance (RT) in the circuit by summing the individual resistances in series:

RT = R1 + R2

Substituting the given values:

RT = 12 Ω + 8 Ω = 20 Ω.

Step 2: Apply Ohm's law to calculate the current (I) in the circuit:

I = V / RT

Substituting the given values:

I = 4 V / 20 Ω = 0.2 A.

Therefore, the current in the circuit is 0.2 A.

Step 3: Calculate the voltage drop across each lamp.

In a series circuit, the voltage is divided among the components based on their resistance. The voltage drop across each lamp can be calculated using the formula:

V1 = I * R1 V2 = I * R2

Substituting the given values:

V1 = 0.2 A * 12 Ω = 2.4 V V2 = 0.2 A * 8 Ω = 1.6 V.

Therefore, the voltage across the first lamp is 2.4 V and the voltage across the second lamp is 1.6 V.

To summarize: - The current in the circuit is 0.2 A. - The voltage across the first lamp is 2.4 V. - The voltage across the second lamp is 1.6 V.

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