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For the circuit shown below let I1, I2 and I3 be the current marked in respective branches, then
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A. I1=0.22A,I2=5.44A,I3=5.66A
B. I1=2A,I2=3A,I3=1A
C. I1=2A,I2=3A,I3=1A
D. I1=2A,I2=3A,I3=1A

Answer
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Hint: The given circuit represents an electrical network. There are three currents in the circuit, namelyI1, I2 and I3. The values of these currents can be found out by Kirchhoff’s Voltage Law or loop theorem. It states that, in a closed loop of an electrical network, the algebraic sum of the potential differences for all the components plus the algebraic sum of emfs is equal to zero.

Complete step by step answer:
The given circuit is a combination of various electric circuit elements (components) and sources of emf and voltages connected in a complicated manner. This is called an electrical network. The emf is positive if we traverse from the negative terminal to the positive terminal inside the cell and the emf is negative if we traverse from the positive terminal to the negative terminal.
In loop ABCFA,
14V+6I110V+4I3=0
This can be also written as
6I1+4I3=24V (1)
In loop FCDEF,
10V+6I12I3=0
This can be also written as
6I12I3=10V (2)
Subtracting equation (1) from equation (2)
6I3=34VI3=5.66A
Substituting the value of I3 in equation (2)
6I12(5.66)=106I1=10+11.336I1=1.33I1=0.22A
Sum of currents flowing toward a junction is equal to the sum of currents flowing away from the junction. Therefore, at junction C
I2+I1=I3I2=I3I1I2=5.660.22I2=5.44A
The values of currents I1, I2 and I3 are 0.22A, 5.44A and 5.66Arespectively.

Hence, option A is the correct answer.

Note: The Kirchhoff’s Voltage Law or loop theorem is only applicable for closed circuits. As the sun of all the potential differences and the emf’s is zero, the Kirchhoff’s Voltage Law is in accordance with the conservation of energy. In that closed loop of electrical network, if the direction of tracing is same as that of the conventional current flow then the potential difference across the resistances is considered as negative otherwise it is positive.