(a) A current of `1 A` flows in a series curcuit containing an electric lamp and a conductor of `5 Omega` when connected to a `10 V` battery. Calculate the resistance of the electric lamp.
(b) Now if a resistance of `10 Omega` is connected in parallel with this series combination, what change (if any) in current flowing through `5 Omega` conductor and potential difference across the lamp will take place ? Give reason.

Explanation

Given current, I = 1A
Resistance of conductor, `R = 5Omega`

Voltage, V=10V
Resistance of lamp, `R_(L)` = ?
Total resistance in the circuit,
`" "R_(T) = (V)/(F) = (10)/(1) = 10 Omega`
`therefore` `" "R_(L) = R_(T) - R = 10 - 5 = 5 Omega`
Potential difference across the lamp
`" "=IR_(L) = 1 xx 5 = 5 V`
When `10Omega` resistance is connected in parallel with total resistance `R_(T) (R_(L) + R = 10 Omega)`, then total resistance R' in the circuit is given by

`(1)/(R') = (1)/(10) + (1)/(R_(T)) = (1)/(10) + (1)/(10) = (2)/(10) = (1)/(5)`
`" "R' = 5 Omega`
Current through the circuit,
`" "I' = (V)/(R') = (10)/(5) = 2A`
Since, `10Omega` and `R_(T) (10Omega)` are in parallel, current through `R_(T)` is `(I')/(2) = (2)/(2) = 1A`
Thus, current through lamp and conductor of `5Omega` in series is 1A. Also PD across lamp
`" "=(I')/(2) xx 5 = 1 xx 5 = 5V`