Thevenin and Norton Theorems for GATE: Dependent Sources, Maximum Power and Solved Examples

Thevenin and Norton theorems for GATE: three ways to find Rth, a solved example with a dependent source, maximum power transfer, common traps and practice questions.

If you had to pick one idea from Network Theory to master before GATE, I'd pick Thevenin's theorem. It turns a messy circuit into one source and one resistor. Once you can do that quickly, a whole family of questions on load current, maximum power and source transformation becomes a two-line job.

The catch is dependent sources. Students who are perfectly comfortable with Thevenin for plain resistor circuits often lose marks the moment a little diamond-shaped source appears. So that's where we'll spend most of our time.

If you've read the convolution post, you've already met the idea behind all of this: linearity. Superposition is the reason these theorems work.

The two theorems in one picture

Thevenin's theorem: any linear circuit, seen from two terminals $a$ and $b$, behaves exactly like a single voltage source $V{th}$ in series with a single resistance $R{th}$.

Norton's theorem: the same circuit also behaves like a current source $IN$ in parallel with a resistance $RN$.

The two are really one result written two ways. They're linked by

where $V{oc}$ is the open-circuit voltage across $a$ and $b$, and $I{sc}$ is the current that flows if you short $a$ to $b$. So $V{th} = V{oc}$ and $IN = I{sc}$.

"Linear" matters here. The circuit can have resistors, independent sources and dependent sources whose value is proportional to some voltage or current. It can't have diodes or anything else whose behaviour changes with the signal level.

Three ways to find Rth

Finding $V{th}$ is usually the easy part: open the terminals and find the voltage. $R{th}$ is where the method depends on the circuit.

Method 1: only independent sources. Switch off every independent source. A voltage source becomes a short circuit and a current source becomes an open circuit. Then combine the resistors as seen from $a$ and $b$.

Method 2: test source. Switch off the independent sources but leave every dependent source in place. Connect a test voltage $Vt$ across $a$ and $b$, find the current $It$ it pushes in, and

Method 3: open and short. Find $V{oc}$ and $I{sc}$ with all sources left as they are, and divide.

Methods 2 and 3 both work with dependent sources. Method 1 never does.

Warm-up: a plain voltage divider

A 12 V source feeds a 6 Ω resistor in series, and a 3 Ω resistor sits across the terminals $a$ and $b$.

With $a$ and $b$ open, the 3 Ω resistor gets its share of the voltage:

Short the source and the two resistors appear in parallel from the terminals:

So the Norton current is $IN = 4 / 2 = 2$ A. You can check it directly: short $a$ to $b$ and the 3 Ω resistor carries nothing, so the current is simply $12 / 6 = 2$ A. It matches.

Worked example with a dependent source

Now the kind of circuit GATE likes. A 10 V source and a 2 Ω resistor feed node $a$. From $a$ to $b$ there's a 4 Ω resistor, with the voltage $Vx$ across it, and a dependent current source that pushes $0.5\,Vx$ amperes into node $a$.

Step 1: open-circuit voltage. With the terminals open, $Vx$ is simply the voltage at $a$. Adding up the currents at node $a$:

The $Vx$ terms on the left cancel, leaving $5 = Vx / 4$, so

Yes, that's twice the source voltage. It isn't a mistake. The dependent source is feeding energy into node $a$, and a circuit with dependent sources is allowed to do this. If you see $V{th}$ larger than every independent source, don't panic and start again.

Step 2: short-circuit current. Short $a$ to $b$. Now $Vx = 0$, so the dependent source produces nothing, and all the current comes through the 2 Ω resistor:

Step 3: divide.

Check with a test source. Short the 10 V source, keep the dependent source, and apply $Vt$ at $a$. Then $Vx = Vt$, and the current the test source must supply is

so $R{th} = Vt / It = 4\ \Omega$. Both methods agree.

Notice what would've happened if you'd switched off the dependent source too. You'd get $2 \parallel 4 = 1.33\ \Omega$, which is exactly the kind of wrong answer the exam will offer you as option (b).

Maximum power transfer

This is where Thevenin earns its keep. Once a circuit is reduced to $V{th}$ and $R{th}$, the power in a load resistor $RL$ is

It is largest when $RL = R{th}$, and then

For our worked example that's $20^2 / (4 \times 4) = 25$ W, delivered to a 4 Ω load. The graph shows how quickly the power falls away on either side.

At maximum power transfer the efficiency is only 50%, since the same current flows through $R{th}$ and $RL$ and they're equal. That's fine for a signal stage or an antenna, but it's the last thing you'd want for a power line.

Thevenin and Norton side by side

Thevenin Norton --------- Source $V{th} = V{oc}$ $IN = I{sc}$ Resistance $R{th}$ in series $RN = R{th}$ in parallel Converting $V{th} = IN R{th}$ $IN = V{th} / R{th}$ Best when the load is in series loads are in parallel

Common traps in GATE questions

These are the mistakes I see most often in class tests.

Switching off the dependent source. Never do this when finding $R{th}$. Only independent sources are switched off. Switching sources off the wrong way. A voltage source becomes a short circuit, and a current source becomes an open circuit, never the other way round. Forgetting the controlling variable changes. When you short the terminals or apply a test source, recalculate $Vx$ or $Ix$ in the new circuit. It's rarely the same as before. Dividing by zero. If the circuit has no independent source, $V{oc}$ and $I{sc}$ are both zero. Use a test source instead. Being surprised by the answer. With dependent sources, $V{th}$ can exceed the supply and $R{th}$ can even be negative. Both can be correct. Maximum power for the wrong resistor. $RL = R{th}$ applies when the load can change. If the load is fixed and the source resistance can be chosen, make the source resistance as small as possible. The direction of $IN$. If the short-circuit current flows out of $a$, through the short and into $b$, draw the Norton arrow pointing towards $a$. Keep the arrow consistent with how you measured $I{sc}$.

Practice questions

Give yourself about two minutes each.

A 24 V source feeds a 4 Ω resistor in series, with a 12 Ω resistor across the terminals. Find $V{th}$, $R{th}$ and $IN$. A Norton equivalent has $IN = 5$ A and $RN = 10\ \Omega$. Find its Thevenin equivalent. A circuit has $V{th} = 12$ V and $R{th} = 3\ \Omega$. What load gives maximum power, and how much power is that? In the worked example, change the dependent source from $0.5\,Vx$ to $1 \cdot Vx$. Find the new $R{th}$. What is the efficiency of a circuit working at maximum power transfer?

Answers

$V{th} = 24 \times 12 / 16 = 18$ V, $R{th} = 4 \parallel 12 = 3\ \Omega$, and $IN = 18 / 3 = 6$ A. Check: a short across the terminals carries $24 / 4 = 6$ A. $V{th} = 5 \times 10 = 50$ V in series with $10\ \Omega$. $RL = 3\ \Omega$, and $P{max} = 12^2 / (4 \times 3) = 12$ W. $R{th} = -4\ \Omega$. With the test source, $It = Vt/2 + Vt/4 - Vt = -0.25\,Vt$. Check with the other method: $V{oc} = -20$ V and $I{sc} = 5$ A, and $-20 / 5 = -4\ \Omega$. 50%. Half the power is lost in $R{th}$.

If question 4 felt strange, that's the point of it. A negative resistance just means the dependent source is supplying more power than the resistors use up. You'll meet it again in oscillators. Tell me in the comments which question took you longest, and I'll go through it in a follow-up.