(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 = R_l

Equivalent resistance R_s

= Resistance of conductor + Resistance of lamp

= 5 + R_l

According to Ohm's law V = IR

Substituting we get, 10 = 1(5+R_l)

R_l = 10 - 5 = 5 Ω

Hence, resistance of the electric lamp = 5 Ω

(b) Total resistance

$\dfrac{1}{R$p} = \dfrac{1}{R1 + 5} + \dfrac{1}{10} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1}{5+5} + \dfrac{1}{10} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1}{10} + \dfrac{1}{10} \\[0.5em] \Rightarrow \dfrac{1}{Rp} = \dfrac{2}{10} \\[0.5em] \Rightarrow Rp = 5 Ω

Hence, Current flowing through the circuit I = $\dfrac{10}{5}$ = 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.