Answer:
3. a) The current in R1 is 0.5 A
The current in R₂ is 0.2[tex]\overline {27}[/tex] 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, [tex]R_{E}[/tex], is given as follows;
[tex]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)}[/tex]
Plugging in the values, we get;
[tex]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[/tex]
The equivalent resistance of the circuit, [tex]R_{E}[/tex] = 10 Ω
The current in R1 = The current in the circuit, [tex]I_E[/tex]= V/[tex]R_{E}[/tex]
∴ 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;
[tex]I_2 = \dfrac{R_{E}}{R2 + R3 + R_{E}} \times I_T[/tex]
Which gives;
[tex]\therefore I_2 = \dfrac{10 \, \Omega}{4 \, \Omega + 8 \, \Omega + 10 \, \Omega} \times 0.5A = \dfrac{5}{22} \, A = 0.2 \overline {27} \, A[/tex]
The current flowing through R₂, I₂ = 0.2[tex]\overline {27}[/tex] A
b) The power dissipated in R₁, P₁ = [tex]I_E^2[/tex] × R₁
∴ The power dissipated in R₁, P₁ = (0.5 A)² × 4 Ω = 0.5 W
