a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. 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 –

a) According to the question, resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are placed in series connection.

We know that the total resistance of the series combination of resistors is given by the algebraic sum of individual resistances. So, according to this:

R = 1 Ω + 2 Ω + 3 Ω = 6 Ω

Thus Total Resistance of series combination  = 6 Ω

b) Let I to be the current flowing through the given circuit

Also,

The emf of the battery is E given by = 12 V

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

We know that, according to Ohm’s law,

E = IR

Upon rearranging, we get –

I = E/R

Upon substituting values in above equation, we get

    \[\]

    \[I=\frac{12}{6}=2A\]

Therefore, the current calculated is 2 A

Consider that the Potential drop across 1 Ω resistor is = V 1

Then again using Ohm’s law, value of V 1 can be obtained :

1 = 2 x 1 = 2 V

Consider that the Potential drop across 2 Ω resistor is= V 2

Similarly, the value of V 2  :

2 = 2 x 2 = 4 V

Consider that the Potential drop across 3 Ω resistor is = V 3

Again, the value of V 3 :

3 = 2 x 3 = 6 V

Therefore, the potential drops across three resistors r 1 = 1 Ω , r 2 = 2 Ω and r 3 = 3 Ω are calculated to be

1 = 2 x 1 = 2 V ; V 2 = 2 x 1 = 4 V ; V 3 = 2 x 1 = 6 V