Question
Question: Find the current in the three resistors in figure. −2=0 ⇒i−i1=0. ⇒2+(i−i1)−2=0 ⇒i−i1=0
Thus i=i1 . −−−−(1)
Considering loop BGFCB:
2−(i−i1)−2+i1−i2=0
⇒(i−i1)+i1−i2=0
But we know that i=i1 , hence the first term gets cancelled.
Thus, i1=i2 −−−−(2)
Considering loop CFEDC, we get
2 + {i_2} - ({i_1} - {i_2}) - 2 = 0 \\\
⇒2i2−i1=0 −−−−(3)
Now compare equation 2 and equation 3, we see that the value is only possible when both i1,i2 are zero. Thus we can also confirm that i = 0.Now lets see the current through resistors:
For R1: current is i−i1 , but from (1) we see the value is zero. Hence current through R1 is zero.
For R2: current is i1−i2=0, but from (2) we see the value is zero. Hence current through R2 is zero. Similarly, for R3 we see that the value of current through the resistance is zero.
Hence no current passes through these three resistances.
Additional information: KVL: The law which deals with the conservation of energy for a closed circuit path. The law states that for a closed loop path the algebraic sum of all the emf around any closed loop in a circuit is equal to zero.
Note: From these types of questions it is to be kept in mind that if equal and opposite emf are applied in a circuit, then no current flows through the circuit. Thus we can now apply this concept to other questions where there may be any number of resistances and emf.