Question

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Answer 1

Answer:

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

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;

$R_{E} = R1 +\dfrac{\left( \dfrac{R5 \cdot Ra}{R5 + Ra} + R4 \right) \times (R2 + R3)}{\left( \dfrac{R5 \cdot Ra}{R5 + Ra} + R4 \right) + (R2 + R3)}$

Plugging in the values, we get;

$R_{E} = 4 +\dfrac{\left( \dfrac{4 \times 4}{4 +4} + 10 \right) \times (4 + 8)}{\left( \dfrac{4 \times 4}{4 +4} + 10 \right) + (4 + 8)} = 10$

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;

$I_2 = \dfrac{R_{E}}{R2 + R3 + R_{E}} \times I_T$

Which gives;

$\therefore I_2 = \dfrac{10 \, \Omega}{4 \, \Omega + 8 \, \Omega + 10 \, \Omega} \times 0.5A = \dfrac{5}{22} \, A = 0.2 \overline {27} \, A$

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