Question

3a

Sthre Gedet Last name only (39 34. (a) Unlows are circled. Pind I,. using path (1111 in) E.TV + YK 10 (b) Find Iz, using junction (1.0., "point") DA, X-s5v 9. (c) Pind Iz. using function (1... "point") - (a) 7 nd&using path Hint: V. - V AV's for (e) Find Yzb -- - Vy, using path - any path from b to I.

Answer 1

Solution to this numerical is solved in paper plz check it

750 1. 102 ZA 55V AER CBA " finching & by taking closed loop - vt I Ri Ja Rat Is R3=0 - 55+ 9 I, - 3420 +7x10=0 - 55ta I-GOATO=0 - Sstota Izo a l=us T I = 5A (1) Finding I2. Taking closed loop -15 +4I2 & 10xtno - 75+4 I 2 A700 I2 = = 123A (1) Finding Is Taking Node B. Iz = 3 +7-4 [I3 = 6A " Taking closed loop EDB - Ezt I x x x 20x3 - Ez +6+60 [E3=66w| undt 3s = 10 [vaga inn - run NAD=

Gustav Kirchhoff’s Voltage Law is the second of his fundamental laws we can use for circuit analysis. His voltage law states that for a closed loop series path the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. This is because a circuit loop is a closed conducting path so no energy is lost.

n other words the algebraic sum of ALL the potential differences around the loop must be equal to zero as: ΣV = 0. Note here that the term “algebraic sum” means to take into account the polarities and signs of the sources and voltage drops around the loop.

This idea by Kirchhoff is commonly known as the Conservation of Energy, as moving around a closed loop, or circuit, you will end up back to where you started in the circuit and therefore back to the same initial potential with no loss of voltage around the loop. Hence any voltage drops around the loop must be equal to any voltage sources met along the way.

So when applying Kirchhoff’s voltage law to a specific circuit element, it is important that we pay special attention to the algebraic signs, (+ and -) of the voltage drops across elements and the emf’s of sources otherwise our calculations may be wrong.

But before we look more closely at Kirchhoff’s voltage law (KVL) lets first understand the voltage drop across a single element such as a resistor.