Respuesta :

Answer:

A = 2.4 A

[tex]R_{eq} = 5 \ \Omega[/tex]

Explanation:

The voltage in the circuit, V = 12 V

The given circuit shows four resistors with R₁ and R₂ arranged in series with both in parallel to R₃ and R₄ which are is series to each other

R₁ = 4 Ω

R₂ = 6 Ω

R₄ = 5 Ω

The voltage across R₃ = 6 V

Voltage across parallel resistors are equal, therefore;

The total voltage across R₃ and R₄ = 12 V

The total voltage across R₁ and R₂ = 12 V

The voltage across R₃ + The voltage across R₄ = 12 V

∴ The voltage across R₄ = 12 V - 6 V = 6 V

The current flowing through  R₄ = 6V/(5 Ω) = 1.2 A

The current flowing through R₃ = The current flowing through R₄ = 1.2 A

The resistor, R₃ = 6 V/1.2 A = 5 Ω

Therefore, we have;

The sum of resistors in series are R₁ + R₂ and R₃ + R₄, which gives;

[tex]R_{series \, 1}[/tex] = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω

[tex]R_{series \, 2}[/tex] = R₃ + R₄ = 5 Ω + 5 Ω = 10 Ω

The sum of the resistors in parallel is given as follows;

[tex]\dfrac{1}{R_{eq}} = \dfrac{1}{R_{series \, 1}} + \dfrac{1}{R_{series \, 2}} = \dfrac{R_{series \, 2} + R_{series \, 1}}{R_{series \, 1} \times R_{series \, 2}}[/tex]

Therefore;

[tex]R_{eq} = \dfrac{R_{series \, 1} \times R_{series \, 2}}{R_{series \, 1} + R_{series \, 2}}[/tex]

Therefore;

[tex]R_{eq} = \dfrac{10\times 10}{10 + 10} \ \Omega = 5 \ \Omega[/tex]

[tex]R_{eq} = 5 \ \Omega[/tex]

The value of the current, A, in the circuit, I = V/[tex]R_{eq}[/tex]

A = I = 12 V/(5 Ω) = 2.4 A

A = 2.4 A

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