An electric lamp of resistance $20\ \Omega$ and a conductor of $4\ \Omega$ resistance are connected in series to a $6 e

Question

An electric lamp of resistance $20\ \Omega$ and a conductor of $4\ \Omega$ resistance are connected in series to a $6	ext{ V}$ 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.

Accepted Answer

Given data:
Resistance of electric lamp ($R1$) = $20\ \Omega$
Resistance of conductor ($R2$) = $4\ \Omega$
Voltage of the battery ($V$) = $6	ext{ V}$ (a) Total resistance of the circuit ($Rs$):
In a series combination, the total resistance is the sum of individual resistances. $Rs = R1 + R2$ $Rs = 20\ \Omega + 4\ \Omega = 24\ \Omega$
Thus, the total resistance of the circuit is $24\ \Omega$. (b) Current through the circuit ($I$):
According to Ohm's law, $I = \frac{V}{Rs}$ $I = \frac{6	ext{ V}}{24\ \Omega} = \frac{1}{4}	ext{ A} = 0.25	ext{ A}$
Thus, the current flowing through the circuit is $0.25	ext{ A}$. (c) Potential difference across the electric lamp ($V1$) and conductor ($V2$):
Potential difference across the lamp ($V1$) = $I 	imes R1$ $V1 = 0.25	ext{ A} 	imes 20\ \Omega = 5	ext{ V}$ 
Potential difference across the conductor ($V2$) = $I 	imes R2$ $V2 = 0.25	ext{ A} 	imes 4\ \Omega = 1	ext{ V}$
Thus, the potential difference across the electric lamp is $5	ext{ V}$ and across the conductor is $1	ext{ V}$.

Electricity

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