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
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 :
V 1 = 2 x 1 = 2 V
Consider that the Potential drop across 2 Ω resistor is= V 2
Similarly, the value of V 2 :
V 2 = 2 x 2 = 4 V
Consider that the Potential drop across 3 Ω resistor is = V 3
Again, the value of V 3 :
V 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
V 1 = 2 x 1 = 2 V ; V 2 = 2 x 1 = 4 V ; V 3 = 2 x 1 = 6 V