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

Explanation

As, `I =E/(R_1 +R_2)`
`1 = 10/(x +5)`
`or `
`:.x = 5Omega` is the resistance of the lamp.
When a resistor of resistace `10 Omega` is connected in parallel with `x + 5 = 10Omega`, the total resistance `R_p` of the arrangement is given by,
`1/R_p = 1/10 + 1/10`
`=1/5`
`:. R_p =5 Omega`
`:.` The current in circuit `= E/(R_p)`
`=10/5`
`=2A`
This current (2A) divides equally between `10 Omega ` and (resistance of lamp + `5 Omega`).

`:.` The current through thet lamp will be again 1A, i.e.,it remains the same.
The potential difference across the lamp will be `1xx x = 1 xx 5 = 5V`
Therefore, the potential difference across the lamp will remain the same.