Physics, 08.11.2019 07:31 xojade

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Answer from: JackMonhen

1) R = 7.62 Ω

I = 3.15 A

I₁ = 2.4 A

I₂ = 0.75 A

2) Rₐ = 11.25 Ω

$I_{p}$ = 2.13 A

Rₓ = 48 Ω

$I_{s}$ = 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,

$I_{p}$ = V/R

= 24 / 11.25

= 2.13 A

Hence, the current through the circuit in parallel connection is, $I_{p}$ = 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

$I_{s}$ = V/R

= 24 / 48

= 0.5 A

Hence, the current through the circuit in parallel connection, $I_{s}$ = 0.5 A

Answer from: Quest

i will answer your question if you answer mine

Answer from: Quest

The answer should be a.

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