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 conductorand potential difference across the lam will take palce ? Give the reason.

A current of 1A flows in a series circuit containing an electric lamp

1.	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 conductorand potential difference across the lam will take palce ? Give the reason.
Answer» Solution : 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.