300+ Network Theorem MCQ – Objective Question Answer for Network Theorem Quiz

By applying KCL

\(\frac{V_P − E}{6} + \frac{V_P}{6}\) – 1 = 0

Or

2 V_P – E = 6

Where

(\(\frac{ − V_P + E}{6}\)) = 2

∴ E – V_P  = 12

Or, V_P  = 18 V

∴ E = 30V.

\(\frac{V_P − 100}{10} + \frac{V_P}{10}\) + 2 = 0

Or, 2V_P – 100 + 20 = 0

∴ V_P  = 80/2 = 40V

∴ R = 20Ω.

For finding V_TH,

V_TH = (4 ×10)/(10 + 10) = 2V

For finding R_TH,

R_TH  = 10 || 10

= (10×10)/(10 + 10) = 5 Ω.

To calculate the Thevenin resistance, all the current sources get open-circuited and voltage sources short-circuited.

(\(\frac{1}{s}\) + 1) || (1 + s) = \(\frac{\left(\frac{1}{s} + 1\right)×(1 + s)}{\left(\frac{1}{s} + 1\right) + (1 + s)}\)

= \(\frac{\frac{1}{s} + 1 + 1 + s}{\frac{1}{s} + 1 + 1 + s}\) = 1
So, R_TH  = 1.

While computing the Thevenin equivalent voltage consisting of both dependent and independent sources, we first find the equivalent voltage called the Thevenin voltage by opening the two terminals.

Then while computing the Thevenin equivalent resistance, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched, and open-circuiting the dependent current sources keeping the independent current sources untouched.

Norton current is obtained by shorting the specified terminals. So, it is the short circuit current. It is not the open-circuit current because if specified terminals get open circuited then the current is equal to zero.

Ideal current sources have infinite internal resistance and hence behave like an open circuit whereas ideal voltage sources have zero internal resistances and hence behave as a short circuit. So, to obtain Norton resistance, all voltage sources are shorted and all current sources are opened.

Norton’s theorem works for only linear circuit elements and not non-linear ones such as BJT, semiconductors, etc.

Norton’s theorem states that a combination of voltage sources, current sources, and resistors is equivalent to a single current source IN and a single parallel resistor RN.

According to Norton’s theorem, Isc is found through the output terminals of a network and not the input terminals.

We can use Norton’s theorem only for linear networks. BJT is a non-linear network hence we cannot apply Norton’s theorem to it.

Shorting all voltage sources and opening all current sources we have:
RN = (3||6) + 10 = 12 ohm.

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:

9I1 − 6I2 = 4

− 6I1 + 16I2 = 6

On solving these equations simultaneously

we get I2 = 0.72A, which is the short circuit current.

Shorting all voltage sources and opening all current sources we have:

RN = (3||6) + 10 = 12 ohm.

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:

9I1 − 6I2 = 4

− 6I1 + 16I2 = 6

On solving these equations simultaneously, we get I2 = 0.72A, which is the short circuit current.

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 the current divider:

I = 0.72 × 12/(12 + 5) = 0.5 A.

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.

The magnitude of the current in Norton’s equivalent circuit is equal to the current passing through the short-circuited terminals that are

I = 20/5 = 4A.

Norton’s resistance is equal to the parallel combination of both the 5Ω and 10Ω resistors is

R = (5×10)/15 = 3.33Ω.

The current passing through the 6Ω resistor and the voltage across it due to Norton’s equivalent circuit is

I = 4×3.33/(6 + 3.33) = 1.43A.

The voltage across the 6Ω resistor is

V = 1.43×6 = 8.58V.

So the current and voltage have the same values both in the original circuit and Norton’s equivalent circuit.