Question

Two resistors of 4·0 Ω and 6·0 Ω are connected

1. in series,
2. in parallel, with a battery of 6·0 V and negligible internal resistance.

For each case draw a circuit diagram and calculate the current through the battery.

Answer 1

(a) Resistors connected in series

Total resistance R = 4 + 6 = 10 Ω, V = 6.0 V

∴ Current I = `"V"/"R"`

= `6.0/10.0`

= 0.6 A

(b) Resistors in parallel `1/"R" = 1/4 + 1/6`

R = `(3 + 2)/12`

R = `12/5`

R = 2.4 Ω

V = 6.0 V

Current I = `"V"/"R"`

= `6/2.4`

= 2.5 A