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Three resistors, R1 = 10 Ω, R2 = 20 Ω and R3 = 30 Ω are connected in series to a 12 V battery. Find: a. The equivalent resistance of the circuit. b. The current in each resistor. c. The voltage across each resistor. d. The power lost in each resistor.

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Explanation:

a) The equivalent resistance for resistors in series is the sum:

R = R₁ + R₂ + R₃

R = 10 Ω + 20 Ω + 30 Ω

R = 60 Ω

b) The resistors are in series, so they have the same current.

I = V/R

I = 12 V / 60 Ω

I = 0.2 A

c) The voltage drop can be found with Ohm's law:

V₁ = I R₁ = (0.2 A) (10 Ω) = 2 V

V₂ = I R₂ = (0.2 A) (20 Ω) = 4 V

V₃ = I R₃ = (0.2 A) (30 Ω) = 6 V

d) The power lost in each resistor is current times voltage drop:

P₁ = I V₁ = (0.2 A) (2 V) = 0.4 W

P₂ = I V₂ = (0.2 A) (4 V) = 0.8 W

P₃ = I V₃ = (0.2 A) (6 V) = 1.2 W

onyebuchinnaji

(a) The equivalent resistance of the circuit is 60 Ω.

(b) The current in each resistor is 0.2 A.

(c) The voltage across each resistor is 2 V, 4 V, and 6 V respectively.

(d) The power lost in each resistor is 0.4 W, 0.8 W and 1.2 W.

Equivalent resistance

The equivalent resistance of the circuit is determined as follows;

Rt = R₁ + R₂ + R₃

Rt = 10 + 20 + 30

Rt = 60 Ω

Current in the resistor

The current in the resistors is calculated as follows;

V = IR

I = V/R

I = 12/60

I = 0.2 A

Voltage across each resistor

V1 =IR1 = 0.2 x 10 = 2 V

V2 = IR2  = 0.2 x 20 = 4 V

V3 = 0.2 x 30 = 6 V

Power lost in each resistor

P1 = I²R1 = (0.2)² x 10 = 0.4 W

P2 = I²R2 = (0.2)² x 20 = 0.8 W

P3 = I²R3 = (0.2)² x 30 = 1.2 W

Learn more about series circuit here:https://brainly.com/question/19865219

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