Example 12.7 An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 12.9). 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.
Resistance of electric lamp = R1 = 20 Ω
Resistance of conductor = R2 = 4 Ω
Potential difference = V = 6 V
Finding total Resistance (R)
Since the resistors are connected in series
R = R1 + R2
R = 20 + 4
R = 24 Ω
(b) Finding current (I)
By ohm’s law,
V = I R
𝑉/𝑅 = I
I = 𝑉/𝑅
I = 6/24
I = 1/4
I = 0.25 A
(c) Finding potential difference across lamp and conductor
We know that,
In series connection
Current remains same across all resistors
Potential difference is different across all resistors
By ohm’s law,
V = I R
Electric lamp (R1 = 20 Ω)
V1 = I R1
V1 = 0.25 × 20
= 25/100 × 20
= 1/4 × 20
= 5 V
Electric lamp (R2 = 4 Ω)
V2 = I R2
V2 = 0.25 × 4
= 25/100 × 4
= 1/4 × 4
= 1 V
Thus,
Total resistance of circuit is 24 Ω
Current flowing the circuit is 0.25 A
Potential difference across
Electric lamp is 5V
Conductor is 1 V

CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo

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