Question

Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series.

a) What is the total resistance of the combination?

b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer 1

a) We know that resistors \( r _1 = 1 Ω , r _2 = 2 Ω \ and\ r _3 = 3 Ω \) are combined in series.

The total resistance of the above series combination can be calculated by the algebraic sum of individual resistances as follows:

Total resistance=1Ω+2Ω+3Ω=6Ω

Thus calculated Total Resistance = 6 Ω

b) Let us consider I to be the current flowing the given circuit

Also,

The emf of the battery is E = 12 V

Total resistance of the circuit ( calculated above ) = R = 6 Ω

Using Ohm’s law, relation for current can be obtained as

\( I = \frac{ E }{ R } \)

Substituting values in the above equation, we get

\( I = \frac{ 12 }{ 6 } = 2 A \)

Therefore, the current calculated is 2 A.

Let the Potential drop across 1 Ω resistor = V1

​The value of V1 can be obtained from Ohm’s law as :

V1 = 2×1 = 2V

Let the Potential drop across 2 Ω resistor = V2

The value of V2 can be obtained from Ohm’s law as :

V2 = 2×2 = 4V

Let the Potential drop across 3 Ω resistor = V3

The value of V 3 can be obtained from Ohm’s law as :

V3 = 2×3 = 6V

Therefore, the potential drops across the given resistors \( r _1 = 1 Ω \), \( r _2 = 2 Ω \) and \( r_ 3 = 3 Ω \) are calculated to be

V1 =2×1=2V

V2 =2×2=4V

V3 =2×3=6V

Total Resistance = 6 Ω, and potential drop across 1Ω, 2Ω and 3Ω is 2V, 4V and 6V respectivily.