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Question: Consider the circuit shown in the figure. The current \(I_3\) is equal to: ![](https://www.vedantu...

Consider the circuit shown in the figure. The current I3I_3 is equal to:

A). 5  A5\;A
B). 3  A3\;A
C). −3  A-3\;A
D). −56  A-\dfrac{5}{6}\;A

Explanation

Solution

Use Kirchhoff’s Current Law to obtain an expression for I3I_3 in terms of the currents flowing across the 8  V8\;V and 12  V12\;V supply voltages. Then, divide the circuit into two independent loops and apply Kirchhoff’s Voltage Law to each to arrive at the magnitude of currents across the before mentioned supply voltages. Use Ohm’s Law to find the voltage drops across the resistance elements. Then plug the values of these currents into the Current law expression to arrive at an appropriate value for I3I_3.
Formula Used:
Kirchhoff’s Current Law (KCL): Ientering+Iexiting=0I_{entering} + I_{exiting} = 0
Kirchhoff’s Voltage Law (KVL): ΣV=0\Sigma V =0
Ohm’s law: V=IRV=IR

Complete step-by-step solution:
The only tools we need to solve this problem are Kirchhoff’s Current Law and Kirchhoff’s Voltage Law.
Kirchhoff’s Current Law (KCL) suggests that the algebraic sum of currents entering a node in a circuit will be equal to the sum of currents leaving the node.
Kirchhoff’s Voltage Law (KVL) suggests that the sum of all voltages around any closed loop in a circuit must be equal to zero.

From the above diagram, we see that we can apply KCL to the node in the lower part of the circuit that is in between the 8  V8\;V and 12  V12\;V batteries.
Applying KCL at this node, we get:
I3=I1+I2I_3 = I_1 +I_2
We thus need to find I1I_1 and I2I_2
Now, our circuit can be deconstructed into two loops as shown in the figure. Since our circuit consists of two independent closed loops, let us apply KVL to each and see how it goes from there.
In Loop 1, the 28Ω28\Omega resistor is powered by the current I1I_1 from the 8  V8\;V supply.
Therefore, the voltage drop across this resistor will be V=RI=28I1V = RI = 28I_1
Now, applying KVL to Loop 1, we get:
28I1+6+8=0⇒28I1=−6−8=−1428I_1 + 6 +8 = 0 \Rightarrow 28I_1 = -6-8 = -14
⇒I1=−1428=−12  A\Rightarrow I_1 = \dfrac{-14}{28} = -\dfrac{1}{2}\;A
Similarly, in Loop 2, the 54Ω54\Omega resistor is powered by the current I2I_2 from the 12  V12\;V supply.
Therefore, the voltage drop across this resistor will be V=RI=54I2V = RI = 54I_2
Now, applying KVL to Loop 2, we get:
54I2+6+12=0⇒54I2=−6−12=−1854I_2 + 6 +12 = 0 \Rightarrow 54I_2 = -6-12 = -18
⇒I2=−1854=−13  A\Rightarrow I_2 = \dfrac{-18}{54} = -\dfrac{1}{3}\;A
From the KCL equation, and the value of I1I_1 and I2I_2 that we found, we can now calculate I3I_3:
I3=I1+I2=−12+(−13)=−3−26=−56  AI_3 = I_1 +I_2 = -\dfrac{1}{2} + \left(-\dfrac{1}{3}\right) = \dfrac{-3-2}{6} = -\dfrac{5}{6}\;A. Therefore, the correct choice would be D. −56  A-\dfrac{5}{6}\;A

Note: Though we looked at KCL and KVL from a quantitative perspective, it is important to understand what they mean in a physical sense.
KCL signifies conservation of charge since the law basically suggests that the sum of currents entering a node must be equal to the sum of currents leaving the node, which means that electric charges are neither ambiguously lost nor mysteriously added but remains the same in an isolated system.
KVL signifies conservation of energy since the total energy in a system remains constant, though it may be transferred between components of the system in the form of electric potential and current.