# Circuit Theory | 10+ Important Mathematical Problems Circuit Theory Numerical Problems

Image Credit: Gabriela P.Printed Circuit Boards in a sheetCC BY 4.0

### [Specially picked questions for GATE, JEE, NEET]

In the Circuit theory series, we have come across some fundamental yet essential rules, formulas, and methods. Let us find out some applications of them and understand them more clearly. The problems will be mainly on – KCL, KVL, Thevenin’s theorem, Norton’s theorem, Superposition theorem, Maximum Power Transfer Theorem.

## Circuit Theory: 1. Find out the maximum power which can be transferred to the load RL for the below-given circuit. Apply required theorems of Circuit Theory. Circuit Theory Problems, Image – 1
• Solution: Remove the load resistor from the circuitry and voltage source to find out Equivalent Resistance.

So, the resistance or the impedance (AC Circuit) of the circuit through the open terminal:

ZTH = 2 || j2 = (2 x j2) / (2 +j2) = j2 / (1 +j)

Or, ZTH = 2 ∠90o / √2 ∠45o

Or, ZTH = √2 ∠45o

Now, we will calculate the current through the j2 ohms resistor.

I = 4 ∠0o / (2 +j2)

Or, I = 2 / (1+j) = √2 ∠ – 45o

The Thevenin’s equivalent voltage comes as VTH = I * j2.

Or, VTH = 2√2 ∠45o V Circuit Theory Problems, Image – 2

Now we can redraw the circuit in Thevenin’s equivalent circuit.

Now, from power transfer theorem, RL = | ZTH| = √2 ohm for full power.

Now, The current through the load IL = VTH / (RTH + RL)

Or, IL = 2√2 ∠45o / (√2 + √2 ∠45o)

OR, IL = 2 ∠45o / (1 + 1∠45o)

OR, IL = 2 ∠45o / [ 1 + (1 + √2) + (j / √2)]

OR, IL = 1.08 ∠22.4o A

|IL| = 1.08 So, the maximum power is: |IL2| RL = (1.08 x 1.08) x √2 = 1.65 W.

## Circuit Theory: 2. Find out Norton’s equivalent resistance at terminal AB, for the below-given circuit. Circuit Theory Problems, Image – 3
• Solution: At first, we will apply a voltage source at the open circuit at the AB terminal. We name it VDC and assume IDC flows from it.

Now, we apply Kirchhoff’s Current Law to do nodal analysis at node a. We can write, Circuit Theory Problems, Image – 4

(Vdc – 4I) / 2 + (Vdc / 2) + (Vdc / 4) = Idc

Here, I = Vdc / 4

Or, 4I = Vdc

Again, (Vdc – Vdc) / 2 + Vdc / 2 + Vdc / 4 = Idc

Or, 3Vdc / 4 = Idc

And, Vdc / Idc = RN

Or, RN = 4/3 = 1.33 ohm.

So, the Norton’s Equivalent Resistance is 1.33 ohms.

## Circuit Theory: 3. Find out the value of R1 in Delta equivalent circuit of the given star connected network. Circuit Theory Problems, Image – 5
• Solution: This problem can be solved easily, using the star’s conversion formula to delta connection. Circuit Theory Problems, Image – 6

Let us assume, that Ra = 5 ohms, Rb = 7.5 ohms, and Rc = 3 ohms.

Now, applying the formula,

R1 = Ra + Rc + (Ra * Rc / Rb)

Or, R1 = 5 + 3 + (5 x 3) / 7.5

Or, R1 = 5 + 3 + 2 = 10 ohms.

So, the R1 Delta Equivalent resistance is: 10 ohms.

## Circuit Theory: 4. Find out the current flowing through the R2 resistor for the circuit given below. Circuit Theory Problems, Image – 7
• Solution: We have to use the idea of source transformation and Kirchhoff’s Voltage Law to find out the answer.

Let us assume ‘I’ Ampere current flows through the R2 (1 kilo-ohm resistor). We can say current through 2-kilo ohm resistance will be (10 – I) Ampere (As current from 10 A source will be 10 A). Similarly, current from 2 A quotation will be 2 A and thus current through 4-kilo ohm resistance will be (I – 2) Ampere.

Now, we apply Kirchhoff’s voltage law in the loop. We can write Circuit Theory Problems, Image – 8

I x 1 + (I – 2) x 4 + 3 x I – 2 x (10 – I) = 0

Or, 10I – 8 – 20 = 0

Or, I = 28/10

Or, I = 2.8 mA

So, the current through the R2 resistor is 2.8 mA.

## Circuit Theory: 5. If the equivalent resistance for the infinite parallel ladder given in the below image is Req, calculate Req / R. Also find the value of Req when R = 1 ohm. Circuit Theory Problems, Image – 9
• Solution: To solve the problem, we must know the equivalent resistance of the infinite parallel ladder. It is given by RE = R x (1 + √5)/2.

So, we can replace the circuit in the following one. Circuit Theory Problems, Image – 10

The equivalent resistance comes here: Req = R + RE = R + 1.618R

Or, Req / R = 2.618

And when R = 1 ohm, Req = 2.618 x 1 = 2.618 ohm.

## Circuit Theory: 6. A source voltage supplies voltage, Vs(t) = V Cos100πt. The source has an internal resistance of (4+j3) ohm. Find out the resistance of a purely resistive load, for transferring maximum power.

• Solution: We know that the power transmitted for a purely resistive circuit is the average power transferred.

So, RL = √ (Rs2 + Xs2)

Or, RL = √ (42 + 32)

Or, RL = 5 ohm.

So, the load will be of 5 ohms.

## Circuit Theory: 7.  Find out the Thevenin’s equivalent impedance between node 1 and 2 for the given circuit. Circuit Theory Problems, Image – 11
• Solution: To find the Thevenin’s equivalent impedance, we need to connect a voltage source of 1 volt in the place of node 1 and 2. Then we will calculate the current value.

So, ZTH = 1 / ITH

ZTH is the desired resistance we have to find. ITH is the current flowing due to the voltage source.

Now applying Kirchhoff’s Current law at node B,

iAB + 99ib – ITH =0

Or, iAB + 99ib = ITH ——- (i)

Applying KCL at node A,

ib – iA – iAB = 0

or, ib = iA + iAB ——- (ii) Circuit Theory Problems, Image – 12

From equation (i) and (ii) we can write,

ib – iA + 99ib = ITH

Or, 100ib – iA = ITH ——- (iii)

Now, we apply Kirchhoff’s Voltage law at the outer loop,

10 x 103ib = 1

Or, ib = 10-4 A.

And also,

10 x 103ib = – 100iA

Or, iA = – 100iA

From equation (iii), we can write,

100iA + 100ib = ITH

Or, ITH = 200ib

Or, ITH = 200 x 10-4 = 0.02

So, ZTH = 1 / ITH = 1 / 0.02 = 50 ohms.

S, the impedance in-between node 1 and 2 is 50 ohms.

## Circuit Theory: 8. A complex circuit is given below. Let us assume that both the voltage source of the circuit is in phase with each other. Now, the circuit is divided virtually in two-part A and B by the dotted lines. Calculate R’s value in this circuit for which maximum power is transferred from Part A to Part B. Circuit Theory Problems, Image – 13
• Solution: The problem can be solved in a few steps.

First, we find the current ‘i’ through R.

Or, i = (7 / (2 – R) A

Next, current through the 3V source,

i1 = i – (3 / -j)

Or, i1 = i – 3j

Then, we calculate the power transferred from Circuit B to A.

P = i2R + i1 x 3

Or, P = [7 / (2 – R)]2 x R + [7 / (2 – R)] x 3 —- (i)

Now, condition for transferring the maximum power is, dP / dR = 0.

So, differentiating equation (i) with respect to R, we can write:

[7 / (2 – R)]2 + 98R/ (2 + R)2 – 21/ (2 + R)2 = 0

Or, 49 x (2 + R) – 98R – 21 x (2 + R)2 = 0

Or, 98 + 42 = 49R + 21R

Or, R = 56 / 70 = 0.8 ohm

So, the R value for maximum power transfer from A to B is 0.8 ohm.

## Circuit Theory: 9. Find out the value of the resistance for maximum power transferring. Also, find out the maximum delivered power. Circuit Theory Problems, Image – 14
• Solution: At the first step, remove the load and calculate the Thevenin’s Resistance.

VTH = V * R2 / (R1 + R2)

Or, VTH = 100 * 20 / (20 +30)

Or, VTH = 4 V

The resistors are parallelly connected.

So, RTH = R1 || R2

Or, RTH = 20 || 30

Or, RTH = 20 * 30 / (20 + 30)

Or, RTH = 12 Ohms

Now the circuit is redrawn using the equivalent values. For maximum power transfer, RL = RTH = 12 ohms.

Maximum power PMAX = VTH2 / 4 RTH.

Or, PMAX = 1002 / (4 × 12)

Or, PMAX = 10000 / 48

Or, PMAX = 208.33 Watts

So, the maximum delivered power was 208.33 watts.

## Circuit Theory: 10. Calculate the load for maximum power transferring.Find out the transferred power also. Circuit Theory Problems, Image – 15
• Solution:

At the first step, remove the load, and calculate the Thevenin’s voltage now. Circuit Theory Problems, Image – 16

So, VAB = VA – VB

VA comes as: VA = V * R2 / (R1 + R2)

Or, VA = 60 * 40 / (30 + 40)

Or, VA = 34.28 v

VB comes as:

VB = V * R4 / (R3 + R4)

Or, VB = 60 * 10 / (10 + 20)

Or, VB = 20 v

So, VAB = VA – VB

Or, VAB = 34.28 – 20 = 14.28 v

In the next step, calculation of resistance. As the rule says, remove the voltage and short circuit the connection. Circuit Theory Problems, Image – 17

RTH = RAB = [{R1R2 / (R1 + R2)} + {R3R4 / (R3 + R4)}]

OR, RTH = [{30 × 40 / (30 + 40)} + {20 × 10 / (20 + 10)}]

OR, RTH = 23.809 ohms Circuit Theory Problems, Image – 18

Now, draw the connection again with the calculated values. For maximum power transfer, RL = RTH = 23.809 ohms.

The load value will be = 23.809 ohms.

Maximum power is PMAX = VTH2 / 4 RTH.

Or, PMAX = 14.282 / (4 × 23.809)

Or, PMAX = 203.9184 / 95.236

Or, PMAX = 2.14 Watts

So, the maximum delivered power was 2.14 watts.

About Sudipta Roy I am an electronics enthusiast and currently devoted towards the field of Electronics and Communications.
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