This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Norton’s Theorem”.
1. The Norton current is the_______
a) Short circuit current
b) Open circuit current
c) Open circuit and short circuit current
d) Neither open circuit nor short circuit current
Explanation: Norton current is the short circuit current. It is the current through the specified load resistance. It is not the open circuit current because open circuit current is equal to zero.
2. Norton resistance is found by?
a) Shorting all voltage sources
b) Opening all current sources
c) Shorting all voltage sources and opening all current sources
d) Opening all voltage sources and shorting all current sources
Explanation: Current sources have infinite internal resistance hence behave like an open circuit whereas ideal voltage sources have 0 internal resistances hence behave as a short circuit.
3. Norton’s theorem is true for __________
a) Linear networks
b) Non-Linear networks
c) Both linear networks and nonlinear networks
d) Neither linear networks nor non-linear networks
Explanation: Norton’s theorem works for only linear circuit elements and not non-linear ones such as BJT, semiconductors etc.
4. In Norton’s theorem Isc is__________
a) Sum of two current sources
b) A single current source
c) Infinite current sources
Explanation: Norton’s theorem states that a combination of voltage sources, current sources and resistors is equivalent to a single current source Ith and a single parallel resistor R.
5. Isc is found across the ____________ terminals of the network.
c) Neither input nor output
d) Either input or output
Explanation: According to Norton’s theorem, Isc is found through the output terminals of a network and not the input terminals.
6. Can we use Norton’s theorem on a circuit containing a BJT?
c) Depends on the BJT
d) Insufficient data provided
Explanation: We can use Norton’s theorem only for linear networks. BJT is a non-linear network hence we cannot apply Norton’s theorem for it.
Explanation: Shorting all voltage sources and opening all current sources we have:
RN=(3||6)+10= 12 ohm.
Explanation: Since the 5 ohm is the load resistance, we short it and find the resistance through the short.
If we apply source transformation between the 6 ohm resistor and the 1A source, we get a 6V source in series with a 6 ohm resistor. Now we have two meshes. Let us consider I1 flowing in the first mesh and I2 flowing in the second mesh.
The mesh equations are:
On solving these equations simultaneously, we get I2=0.72A, which is the short circuit current.
Explanation: From Q8 and Q7 we have found the values of the Isc and RN respectively.
Connecting the current source in parallel to RN which is in turn connected in parallel to the load resistance=5ohm, we get Norton’s equivalent circuit.
Using current divider: I=0.72*12/(12+5)= 0.5 A.
10. Which of the following is also known as the dual of Norton’s theorem?
a) Thevenin’s theorem
b) Superposition theorem
c) Maximum power transfer theorem
d) Millman’s theorem
Explanation: Thevenin’s theorem is also known as the dual of Norton’s theorem because in Norton’s theorem we find short circuit current which is the dual of open circuit voltage-what we find in Thevenin’s theorem.
Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.
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