NCERT · Class 10 · Science · ElectricityAn electric lamp of resistance $20\ \Omega$ and a conductor of $4\ \Omega$ resistance are connected in series to a $6\text{ V}$ battery. Calculate: (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.

Given data:\nResistance of electric lamp ($R_1$) = $20\ \Omega$\nResistance of conductor ($R_2$) = $4\ \Omega$\nVoltage of the battery ($V$) = $6\text{ V}$

(a) Total resistance of the circuit ($R_s$):\nIn a series combination, the total resistance is the sum of individual resistances. $R_s = R_1 + R_2$ $R_s = 20\ \Omega + 4\ \Omega = 24\ \Omega$\nThus, the total resistance of the circuit is $24\ \Omega$.

(b) Current through the circuit ($I$):\nAccording to Ohm's law, $I = \frac{V}{R_s}$ $I = \frac{6\text{ V}}{24\ \Omega} = \frac{1}{4}\text{ A} = 0.25\text{ A}$\nThus, the current flowing through the circuit is $0.25\text{ A}$.

(c) Potential difference across the electric lamp ($V_1$) and conductor ($V_2$):\nPotential difference across the lamp ($V_1$) = $I \times R_1$ $V_1 = 0.25\text{ A} \times 20\ \Omega = 5\text{ V}$ \nPotential difference across the conductor ($V_2$) = $I \times R_2$ $V_2 = 0.25\text{ A} \times 4\ \Omega = 1\text{ V}$\nThus, the potential difference across the electric lamp is $5\text{ V}$ and across the conductor is $1\text{ V}$.