Problem 31: The parallel circuit depicts three resistors connected to a voltage source.
The voltage source (AVtot) is a 110.-V source and the resistor values are
15.4 QR1), 21.9 D (R2) and 11.7 Q (R3).
b. Determine the current in each branch resistor.
c. Determine the total current in the circuit.

Bi. Current in 15.4 Ω (R₁) is 7.14 A.

Bii. Current in 21.9 Ω (R₂) is 5.02 A.

Biii. Current in 11.7 Ω (R₃) is 9.40 A.

C. Total current in the circuit is 21.56 A.

Explanation:

Bi. Determination of the current in 15.4 Ω (R₁)

Voltage (V) = 110 V

Resistance (R₁) = 15.4 Ω

Current (I₁) =?

V = I₁R₁

110 = I₁ × 15.4

Divide both side by 15.4

I₁ = 110 / 15.4

I₁ = 7.14 A

Therefore, the current in 15.4 Ω (R₁) is 7.14 A.

Bii. Determination of the current in 21.9 Ω (R₂)

Voltage (V) = 110 V

Resistance (R₂) = 21.9 Ω

Current (I₂) =?

V = I₂R₂

110 = I₂ × 21.9

Divide both side by 21.9

I₂ = 110 / 21.9

I₂ = 5.02 A

Therefore, the current in 21.9 Ω (R₂) is 5.02 A

Biii. Determination of the current in 11.7 Ω (R₃)

Voltage (V) = 110 V

Resistance (R₃) = 11.7 Ω

Current (I₃) =?

V = I₃R₃

110 = I₃ × 11.7

Divide both side by 11.7

I₃ = 110 / 11.7

I₃ = 9.40 A

Therefore, the current in 11.7 Ω (R₃) is 9.40 A.

C. Determination of the total current.

Current 1 (I₁) = 7.14 A

Current 2 (I₂) = 5.02 A

Current 3 (I₃) = 9.40 A

Total current (Iₜ) =?

Iₜ = I₁ + I₂ + I₃

Iₜ = 7.14 + 5.02 + 9.40

Iₜ = 21.56 A

Therefore, the total current in the circuit is 21.56 A

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation: