Bismill

__a__hirrohm__a__nirroh__i__m# Kirchhoff's voltage law

**KVL (Kirchhoff's Voltage Law)**reads:

In any closed circuit the amount of electromotive force and potential difference is zero, or the sum of electrical voltage in a circuit equals to the sum product of current and resistance

written by equation

Σ E = Σ (I × R)

Procedure that followed in determining the circuit equation is:

- Pull an arrow from negative pole (-) to positive pole (+) of the voltage source, to each battery or generator
- Determine the direction of current flow, and show the direction of the circuit by an arrows. If specified the wrong direction, then the calculation of current value becomes negative, it indicates that the current is flowing in the opposite direction
- Beside each resistance describing potential difference pointing arrow in the opposite direction of current flow that flows through the resistance
- Surround closed loop or circuit starting from any point and end at that point anyway
- Write down Kirchoff voltage equations for the loop
- For potential arrows and potential difference arrow in a clockwise direction, given a positive sign (+) and the opposite was given a negative sign (-)

Example question 1:

Known to the circuit as shown below, where the value of E2> E1, determine the equation given in the circuit

completion:

Based on the procedure KVL, drawing the circuit becomes

Circuit equations are written with

- I R1 - E1 - I R2 - R3 + I E2 = 0

E2 - E1 = I R1 + I R2 + I R3

Example question 2:

Using an example about an image, if known value of E2 = 6 V, E1 = 4 V, R1 = 10 Ω, R2 = 8 Ω, and R3 = 12 Ω, what is the current flowing in the circuit?

completion:

From the equation obtained, we write

E2 - E1 = I R1 + I R2 + I R3

E2 - E1 = I (R1 + R2 + R3)

I = (E2 - E1) / (R1 + R2 + R3)

I = (6-4) V / (10 + 8 + 12) Ω

I = 2 / 30 A

I = 0.067 A

I = 67 mA

So we get the current flowing in the circuit by 67 mA.

Easily explained all the details.

ReplyDeletejobs in new zealand