Question 8(iii)
(a) A current of 1 A flows in a series circuit having an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp.
(b) Now, if a resistance of 10 Ω is connected in parallel with this series combination, then what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.
(a) Given,
Current I = 1 A
Voltage V = 10 V
Resistance of conductor = 5 Ω
Resistance of lamp = Rl
Equivalent resistance Rs
= Resistance of conductor + Resistance of lamp
= 5 + Rl
According to Ohm's law V = IR
Substituting we get, 10 = 1(5+Rl)
Rl = 10 - 5 = 5 Ω
Hence, resistance of the electric lamp = 5 Ω
(b) Total resistance
Hence, Current flowing through the circuit I = = 2A
Thus, 1A current flows through 10 Ω resistor and 1A flows through lamp and conductor, hence, there is no change in current flowing through conductor. Also there is no change in potential difference across the lamp.