An electric lamp, whose resistance is `20 Omega` and a conductor of `4 Omega` resistance are connected to a `6 V` battery as shown in (Fig. 3.18) Calculate.
(a) the total resistance of the circuit,
(b) the current through the circuit, and
( c) the potential difference across the electric lamp and the conductor.
.

Answer

(a) `24 Omega`
(b) `0.25 A`
(c ) `5 V, 1 V`.

Explanation

Since the lamp (of resistance `R_1 = 20 Omega`) and conductor (of resistance `R_2 = 4 Omega`) are in series,
(a) total resistance in the circuit, `R_s = 20 Omega + 4 Omega = 24 Omega`
(b) current through the circuit, `I = (V)/(R_s) = (6 V)/(24 Omega) = 0.25 A`
( c) pd across the lamp, `V_1 = IR_1 = 0.25 A xx 20 Omega = 5 V`
pd across the conductor, `V_2 = IR_2 = 0.25 A xx 4 Omega = 1 V`.