Question

Two resistors of 2.0 Ω and 3.0 Ω are connected (a) in series (b) 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) In Series -

 

R₁ = 2 Ω

R₂ = 3 Ω

R = R₁ + R₂ = 2 + 3 = 5 Ω

V = 6 V 

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

I `= 6/5`

I = 1.2 A

(b) In Parallel -

 

R₁ and R₂ are connected in parallel 

`1/"R" = 1/"R"_1 + 1/"R"_2`

`1/"R" = 1/2+1/3`

`1/"R" = 5/6` 

R = `6/5` = 1.2 Ω

V = 6V 

I `= "V"/"R"`

`= 6/1.2`

= 5 A