Question

Draw a schematic diagram of a circuit consisting of a battery of six 2V cells, a 6 Ω resistor, a 12 Ω resistor, and a 18 Ω resistor and a plug key all connected in series. Calulate the following (when key is closed):

1. Electric current flowing in the circuit.
2. Potential difference across 18 Ω resistor.
3. Electric power consumed in 18 Ω resistor.

Answer 1

i. Total resistance = R₁ + R₂ + R₃

= 6 + 12 + 18

= 36 Ω

V = 6 × 2

V = 12

Using Ohm's law 

I = `V/R`

= `12/36`

= `1/3`

= 0.33

ii. Ohm's Law determines the potential difference (V₁₈) across the 18 Ω resistor:

V₁₈ = I × R₁₈

= 0.33 × 18

= 5.94

iii. The electric power (P) consumed in the 18 n resistor is given by

P = I² × R

= 0.332 × 18

= 1.96 Watt