Resistors of 4 Ω, 6 Ω and 8 Ω are connected in series across 24 V. Find total resistance and current.

Solution & explanation

1. Given: R₁ = 4 Ω; R₂ = 6 Ω; R₃ = 8 Ω; V = 24 V.

2. In series, resistances add.

   $$R=4+6+8$$

3. Add 4 and 6, then add 8.

   $$R=8+10=18$$

4. Rearrange Ohm’s law V = IR.

   $$I=\frac{V}{R}$$

5. Insert total voltage and resistance.

   $$I=\frac{24}{18}$$

6. Divide numerator and denominator by 6.

   $$I=\frac{4}{3}$$

7. State the result with units: R = 18 Ω; I ≈ 1.33 A through every resistor.

R = 18 Ω; I ≈ 1.33 A