This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Kirchhoff’s Current Law”.
Explanation: The current through the 10 ohm resistor=v1/10=2A.Applying KCL at node 1: i5=i10+i2. i2=6-2=4A.
Thus the drop in the 2 ohm resistor= 4×2=8V.
v1=20V; hence v2=20-v across 2 ohm resistor=20-8=12V
v2=v since they are connected in parallel.
Explanation: KCl states that the total current leaving the junction is equal to the current entering it. In this case, the current entering the junction is 5A+10A=15A.
Explanation: Using current divider rule the current through the 20 ohm resistor= (total current in the circuit x resistance of the other branch)/total resistance= 1×10/30=3.33A.
Explanation: According to KCl, I1+I2+I3=0. Hence I3=-(I1+I2)=-5A.
Explanation: At junction a: i1-i3-i2=0. i2=2A.
At junction b: i4+i2-i6=0. i4=-1A.
At junction c: i3-i5+i4=0. i5=2A.
6. What is the value of current if a 50C charge flows in a conductor over a period of 5 seconds?
Explanation: Current=Charge/Time. Here charge = 50c and time = 5seconds, so current = 50/5 = 10A.
7. KCL deals with the conservation of?
c) Potential Energy
Explanation: KCL states that the amount of charge entering a junction is equal to the amount of charge leaving it, hence it is the conservation of charge.
8. KCL is applied at _________
c) Both loop and node
d) Neither loop nor node
Explanation: KCL states that the amount of charge leaving a node is equal to the amount of charge entering it, hence it is applied at nodes.
9. KCL can be applied for __________
a) Planar networks
b) Non-planar networks
c) Both planar and non-planar
d) Neither planar nor non-planar
Explanation: KCL is applied for different nodes of a network whether it is planar or non-planar.
Explanation: At the junction, I-2+3-4-5=0. Hence I=8A.
Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.
To practice all areas of Basic Electrical Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.