
Find the current in the three resistors in figure.
Answer
502.2k+ views
Hint:To solve circuit-based questions we have to know the concepts and applications of KVL and KCL. Name the ends of the circuits as ABCDEFGH. Let us consider a current $i$ coming out from the lower left most cell of the circuit AB and then dividing into ${i_1}$ and$i - {i_1}$ at point G. Now let ${i_1}$ be again divided into part ${i_1}$ and ${i_1} - {i_2}$ at the point F.
Complete step by step answer:
Considering loop AHGBA:
$ 2 + (i - {i_1}) - 2 = 0 \\
\Rightarrow i - {i_1} = 0. \\
\Rightarrow 2 + (i - {i_1}) - 2 = 0 \\
\Rightarrow i - {i_1} = 0$
Thus $i = {i_1}$ . $ - - - - (1)$
Considering loop BGFCB:
$2 - (i - {i_1}) - 2 + {i_1} - {i_2} = 0$
$\Rightarrow (i - {i_1}) + {i_1} - {i_2} = 0 $
But we know that $i = {i_1}$ , hence the first term gets cancelled.
Thus, ${i_1} = {i_2}$ $ - - - - (2)$
Considering loop CFEDC, we get
$2 + {i_2} - ({i_1} - {i_2}) - 2 = 0 \\$
$\Rightarrow 2{i_2} - {i_1} = 0$ $ - - - - (3)$
Now compare equation 2 and equation 3, we see that the value is only possible when both ${i_1},{i_2}$ are zero. Thus we can also confirm that $i{\text{ = 0}}$.Now lets see the current through resistors:
For $R_1$: current is $i - {i_1}$ , but from (1) we see the value is zero. Hence current through $R_1$ is zero.
For $R_2$: current is ${i_1} - {i_2} = 0$, but from (2) we see the value is zero. Hence current through $R_2$ is zero. Similarly, for $R_3$ 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.
Complete step by step answer:
Considering loop AHGBA:
$ 2 + (i - {i_1}) - 2 = 0 \\
\Rightarrow i - {i_1} = 0. \\
\Rightarrow 2 + (i - {i_1}) - 2 = 0 \\
\Rightarrow i - {i_1} = 0$
Thus $i = {i_1}$ . $ - - - - (1)$
Considering loop BGFCB:
$2 - (i - {i_1}) - 2 + {i_1} - {i_2} = 0$
$\Rightarrow (i - {i_1}) + {i_1} - {i_2} = 0 $
But we know that $i = {i_1}$ , hence the first term gets cancelled.
Thus, ${i_1} = {i_2}$ $ - - - - (2)$
Considering loop CFEDC, we get
$2 + {i_2} - ({i_1} - {i_2}) - 2 = 0 \\$
$\Rightarrow 2{i_2} - {i_1} = 0$ $ - - - - (3)$
Now compare equation 2 and equation 3, we see that the value is only possible when both ${i_1},{i_2}$ are zero. Thus we can also confirm that $i{\text{ = 0}}$.Now lets see the current through resistors:
For $R_1$: current is $i - {i_1}$ , but from (1) we see the value is zero. Hence current through $R_1$ is zero.
For $R_2$: current is ${i_1} - {i_2} = 0$, but from (2) we see the value is zero. Hence current through $R_2$ is zero. Similarly, for $R_3$ 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.
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