Question

The series circuit depicts three resistors connected to a voltage

source. The voltage source (AVtot) is a 110-V source and the resistor
values are 7.2 (R1), 6.2 A2 (R2) and 8.6 22 (R3).
b. Determine the current in the circuit.
A
c. Determine the voltage drops across each individual resistor.

Answer 1

B. Current in the circuit is 5.

Ci. Voltage across 7.2 Ω (R₁) is 36 V

Cii. Voltage across 6.2 Ω (R₂) is 31 V

Ciii. Voltage across 8.6 Ω (R₃) is 43 V

Step-by-step explanation:

We'll begin by calculating the number equivalent resistance in the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 7.2 Ω

Resistor 2 (R₂) = 6.2 Ω

Resistor 3 (R₃) = 8.6 Ω

Equivalent Resistance (R) =?

Since the resistors are in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂ + R₃

R = 7.2 + 6.2 + 8.6

R = 22 Ω

B. Determination of the current.

Voltage (V) = 110 V

Resistance (R) = 22 Ω

Current (I) =?

V = IR

110 = I × 22

Divide both side by 22

I = 110 / 22

I = 5 A

Therefore, the current in the circuit is 5.

Ci. Determination of the voltage across 7.2 Ω (R₁)

Resistor 1 (R₁) = 7.2 Ω

Current (I) = 5 A

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 5 × 7.2

V₁ = 36 V

Therefore, the voltage across 7.2 Ω (R₁) is 36 V

Bii. Determination of the voltage across 6.2 Ω (R₂)

Resistor 2 (R₂) = 6.2 Ω

Current (I) = 5 A

Voltage 2 (V₂) =?

V₂ = IR₂

V₂ = 5 × 6.2

V₂ = 31 V

Therefore, the voltage across 6.2 Ω (R₂) is 31 V

Ciii. Determination of the voltage across 8.6 Ω (R₃)

Resistor 3 (R₃) = 8.6 Ω

Current (I) = 5 A

Voltage 3 (V₃) =?

V₃ = IR₃

V₃ = 5 × 8.6

V₃ = 31 V

Therefore, the voltage across 8.6 Ω (R₃) is 43 V