Three resistors (5, 10, and 15 ohms) are connected in series to a 12 V source. All work must be shown to earn credit.

Question

Three resistors (5, 10, and 15 ohms) are connected in series to a 12 V source. All work must be shown to earn credit.

1. Draw the schematic diagram.
2. What is the equivalent resistance of the circuit?
3. What is the current through the battery?
4. What is the current through each resistor?
5. What is the voltage drop across each resistor?
6. How much power is dissipated through each resistor?

Answer 1

ii.

Resistance in series is additive, so their combined resistance is

R = 5 + 10 + 15 = 30 ohms

iii.

The current through the battery is directly proportional to voltage,

and indirectly proportional to resistance

I = V/R = 12/30 = 0.4 A

iv.

The current is common to everything in the circuit,

including the battery and all the resistors.

v.

The voltage drop is proportional to both current and resistance, i.e.,

V = IR

V₁ = 0.4×5 = 2V

V₂ = 0.4×10 = 4V

V₃ = 0.4×15 = 6V

Note that the voltages across all the resistors need to add up to the

voltage across the battery.

vi.

Power is the product of voltage drop and current.

P₁= V₁I = 2×0.4 = 0.8W

P₂ = V₂I = 4×0.4 = 1.6W

P₃ = V₃I = 6×0.4 = 2.4W