An electric lamp, whose resistance is 20 Omega, and a conductor of 4 Omegaresistance are connected to a 6 V battery (see figure). 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.

An electric lamp, whose resistance is 20 Omega, and a conductor of 4 O

1.	An electric lamp, whose resistance is 20 Omega, and a conductor of 4 Omegaresistance are connected to a 6 V battery (see figure). 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.
Answer» Solution :( a) The resistance of electric lamp, `R_1 = 20 Omega` The resistance of the conductor connectedin series,`R_2 = 4Omega` Then the TOTAL resistance in the circuit `R = R_1 + R_2` `R_s = 20 Omega + 4Omega =24 Omega` ( b )The total POTENTIAL difference across the two terminals of the battery `V = 6V` Now by the Ohm.s law, the current through the circuit is given by `I = V/R_a` ` = 6 V/24 Omega` `=0.25 A` ( C )APPLYING the Ohm.s law to the electric lamp and conductor separately, we get potential difference across the electric lamp, `V_1 = 20 Omega xx 0.25 A` `= 5V` and, that across the conductor, `V_2 = 4Omega xx 0.25 A = 1V` Note: Suppose, we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6 V acrosS the battery terminals will cause a current of 0.25 A in the circuit. The resistance R of this equivalent resistor would be `R =V/I` ` = 6V //0.25 A` `=24 Omega` This is the total resistance of the series circuit, it is equal to the sum of the two resistances.