Answer:
1) R = 7.62 Ω
I = 3.15 A
I₁ = 2.4 A
I₂ = 0.75 A
2) Rₐ = 11.25 Ω
[tex]I_{p}[/tex] = 2.13 A
Rₓ = 48 Ω
[tex]I_{s}[/tex] = 0.5 A
Explanation:
1) Given,
The resistor, R₁ = 10 Ω
The resistor, R₂ = 32 Ω
The voltage, V = 24 v
The effective resistance,
R = R₁ R₂ / (R₁ + R₂)
= 10 x 32 / (10 + 32)
= 7.62 Ω
Hence, the effective resistance, R = 7.62 Ω
The total current
I = V / R
= 24 / 7.62
= 3.15 A
Hence, the total current is the circuit, I = 3.15 A
The current through the first resistor,
I₁ = V / R₁
= 24 / 10
= 2.4 A
Hence, the current through the resistor, I₁ = 2.4 A
The current through the resistor, R₂
I₂ = V / R₂
= 24 / 32
= 0.75 A
Hence, the current through the resistor, I₂ = 0.75 A
2) Given,
The resistor, R₁ = 18 Ω
The resistor, R₂ = 30 Ω
The effective resistance in parallel connection,
Rₐ = R₁ R₂ / (R₁ + R₂)
= 18 x 30 / (18 + 30)
= 11.25 Ω
Hence, the effective resistance in parallel connection, Rₐ = 11.25 Ω
The total current in the parallel connection is given by the formula,
[tex]I_{p}[/tex] = V/R
= 24 / 11.25
= 2.13 A
Hence, the current through the circuit in parallel connection is, [tex]I_{p}[/tex] = 2.13 A
Serial connection,
Rₓ = R₁ + R₂
= 18 + 30
= 48 Ω
Hence, the effective resistance in series connection, Rₓ = 48 Ω
Current in series connection
[tex]I_{s}[/tex] = V/R
= 24 / 48
= 0.5 A
Hence, the current through the circuit in parallel connection, [tex]I_{s}[/tex] = 0.5 A
