Question

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Answer 1

3. a) The current in R1 is 0.5 A

The current in R₂ is 0.2
`\overline {27}` A

b) The power dissipated in R₁ is 0.5 W

Step-by-step explanation:

The given circuit parameters are;

The voltage in the circuit = 5 V

The resistances in the circuit are;

R1 = 3 Ω, R2 = 4 Ω, R3 = 8 Ω, R4 = 10 Ω, R5 = 4 Ω, Ra = 4 Ω

3. a) The equivalent resistance of the circuit,
`R_(E)`, is given as follows;

`R_(E) = R1 +(\left( (R5 \cdot Ra)/(R5 + Ra) + R4 \right) * (R2 + R3))/(\left( (R5 \cdot Ra)/(R5 + Ra) + R4 \right) + (R2 + R3))`

Plugging in the values, we get;

`R_(E) = 4 +(\left( (4 * 4)/(4 +4) + 10 \right) * (4 + 8))/(\left( (4 * 4)/(4 +4) + 10 \right) + (4 + 8)) = 10`

The equivalent resistance of the circuit,
`R_(E)` = 10 Ω

The current in R1 = The current in the circuit,
`I_E`= V/
`R_(E)`

∴ I = 5 V/(10 Ω) = 0.5 A

The current in R1 = 0.5A

Let, 'I₂' represent the current flowing through R₂

By the current divider rule, we have;

`I_2 = (R_(E))/(R2 + R3 + R_(E)) * I_T`

Which gives;

`\therefore I_2 = (10 \, \Omega)/(4 \, \Omega + 8 \, \Omega + 10 \, \Omega) * 0.5A = (5)/(22) \, A = 0.2 \overline {27} \, A`

The current flowing through R₂, I₂ = 0.2
`\overline {27}` A

b) The power dissipated in R₁, P₁ =
`I_E^2` × R₁

∴ The power dissipated in R₁, P₁ = (0.5 A)² × 4 Ω = 0.5 W