UNIT 1 | unit 1 dc circuits

BEEE

UNIT 1

DC Circuits

Explain Ohm’s Law.

Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions remain unchanged. Mathematically it can be represented as,

Potential difference ∝ Current

V ∝ I

(When the value of V increases the value of I increases simultaneously)

V = IR ………………………. (eq.1)

Where,

V is Voltage in volts (V)
R is Resistance in ohm (Ω)
I is Current in Ampere (A)

Calculating Various Parameters Using Ohm’s Law

As an equation (eq.1), the Ohm’s Law serves as master for calculating the current (I), when the resistance (R) and the potential difference (V) are known. Similarly, if any two parameters in the equation (eq.1) are known, then the unknown third one can be easily calculated as follows:

To find Voltage(V),

V = IR …………………………… (eq.2)

To find Current(I),

I=VR ……………………………… (eq.3)

To find Resistance(R),

R=VI ……………………………… (eq.4)

Resistance

Resistance is a opposition offered to the current flowing in any electrical circuit. It is measure in terms of ohms, which is symbolized by the Greek letter omega (Ω)

From the definition, it can be said that the unit of electric resistance is volt per ampere.

One unit of resistance is such a resistance which causes 1 ampere current to flow through it when 1 volt potential difference is applied across the resistance.

https://www.electrical4u.com/images/2017/april/1493434445.GIF

The unit of electric resistance is volt per ampere which is called as ohm(Ω) after the name of great German physicist George Simon Ohm.

2. Apply KVL and find I1 and I2.

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First, identify no of loops and assign the direction of current flowing in the loop

Note: no of loops in-circuit = No, of unknown currents = no, of equations in the circuit

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Note: keep loop direction and current direction same ie either clockwise or anticlockwise for all loops I1 I2

Now according to the direction of direction assign signs (+ve to –ve) to the resistors

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Note: voltage sources (V) polarities do not change is constant.

Note: for common resistor between 2 loops appearing in the circuit like R3 give signs according to separate loops as shown

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 When considering only loop no 1 (+ R3 - )

- B

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Now consider diagram A and write equations

Two loops two unknown currents two-equation

Apply KVL for loop ① [B. Diagram ]

(+ to drop -) = - sign and (- to rise +) = + sign

for drop = -sign

for rise = + sign

-() R2 is considered because in R3,2 currents are flowing and and we have taken () because we are considering loop no 1 and current flowing is in the loop no 1

)

Similarly for loop no. 2 currents flowing is resistor R3 it should be )R3

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Consider loop no. 1 apply KVL

- …….①

Consider loop no. 2 apply KVL

-…….②

After solving equation ① and ② we will get branch current and

3. Explain Current division rule.

A parallel circuit acts as a current divider as the current divides in all the branches in a parallel circuit, and the voltage remains the same across them. The current division rule determines the current across the circuit impedance. The current division is explained with the help of the circuit shown below:

The current I has been divided into I1 and I2 into two parallel branches with the resistance R1 and R2 and V is the voltage drop across the resistance R1 and R2.

As we know,

V = IR ……..(1)

Then the equation of the current is written as:

I1 = V/R1 and I2 = V /R2

Let the total resistance of the circuit be R and is given by the equation shown below:

R = R1R2/ R1 + R2

Equation (1) can also be written as:

I = V/R ……….(3)

Now, putting the value of R from the equation (2) in the equation (3) we will get

I = V(R1+R2)/ R1R2 ---------------------------(4)

But

V = I1 R1 = I2 R2 ------------------------------(5)
Putting the value of V = I1R1 from the equation (5) in equation (4), we finally get the equation as:

I = I1R1(R1+R2)/ R1R2 = I1(R1+R2)/ R2---------------------(6)

And now considering V = I2R2 the equation will be:

I = I2R2(R1+R2)/ R1R2 = I1/R1 (R1+R2) ------------------------------(7)

Thus, from the equation (6) and (7) the value of the current I1 and I2 respectively is given by the equation below:

I1 + I . R2/ R1 + R2 and I2 = I . R1 / R1 + R2

Thus, in the current division rule, it is said that the current in any of the parallel branches is equal to the ratio of opposite branch resistance to the total resistance, multiplied by the total current.

4. Convert Delta to Star and find the required expression.

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Equivalent resistance between ① and ②

DeltaStar

= R12// (R23 + R13) =R1 + R2

= = R1 + R2

Here let R = R12 + R23 + R13

Similarly, we can find Req. Between 2 and 3

= R2 + R3

Similarly, we can find req. Between 1 and 3

R1 + R3

Now the 3 equations after equating L.H.S. And R.H.S

R1 + R2 = …….①

R2 + R3 = ……②

R1 + R3 = …..③

Now subtract ② and ① on L.H.S. And R.H.S

R2+ R3 – R1 – R2 =

R3 – R1 = …..④

Now add equation ④ and ③

R3 – R1 + R1 + R3 =

2R3 =

Similarly, R1 =

And R2 = R23 R12/R where R = R12 + R23 + R13

i.e. star equivalent from delta network is the ratio of the product of adjacent branches in the delta to the addition of all branches in the delta.

5. Explain Voltage Division Rule.

The voltage division rule can be understood by considering a series circuit shown below. In a series circuit, voltage is divided, whereas the current remains the same.

Let us consider a voltage source E with the resistance r1 and r2 connected in series across it.

As we know,

I = V/R or we can say I = E/R

Therefore, the current (i) in the loop ABCD will be:

i = E / r1 + r2 -------------------------(8)

And r1 = ir1

By putting the value of I from equation (8) in equation (9) the voltage across the resistance r1 and r2 respectively are given by the equation shown below as:

E1 = Er1/r1 + r2 and E2 = E r2/ r1 + r2

Thus, the voltage across a resistor in a series circuit is equal to the value of that resistor times the total impressed voltage across the series elements divided by the total resistance of the series elements.

6. Convert Star to Delta and find the required expression.

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