Question

5) Given the following circuit: 4.00 12 8.00 22 3.00 A 24.0 V 6.00 2 Www +E V 2.00 12 a) Find the current I through the 2.00 22 resistor and the voltage V of the battery to the left of this resistor b) Calculate the power delivered to all elements in the circuit and show energy is conserved.

I have trouble understanding the Kirchoff Loop used here, could this be explained in detail with the formulas used if any clearly illustrated? Much will be appreciated, thanks!

Answer 1

Salt ID I, tl, = I fat junction ero) for first loop (Ev=o) gyu Iz or + -I 4r + 1,782 +24+ Istberno 77 -2x122 +240 + Iz xberto o for II loop (Euro) - Izaber - 24v- I (22) tv 20 - - Izxb 24 (I1+I₂ ) x 2 tuto I = 3A - Iz X6-24 - (3 + 2) + 2 tu 7 – 6 Iz ~24 - 6-2 Iz tv to be -) – 8 Iz - 30 tuto eq - 3x12 + 24 + 6 Iz 0 =) - 36 of 24+ 6 Iz o. - 12 & 6 Iz 20 I2 = 12 - 2 A 6 It Is - I 2 3+2 = SA ISA Ang Scanned By SelfScan

- by equ- - 8 Iz - 30 tuzo - 8x2 - 30 tund =) - 16-30 tuno VT Ubu Ana Energy conservation Pav.I - IZR Pi 4317 JPY ZA > P3 I-SA 22 to < Po uou PS avea consumed ave & supplices for the entine circuit Ew=0 P = I, 2xR = 944= 36W P2 = (3) 7 8 = 72W Pg = (2,²x6 Py = VRI- 24x2=480 2uw = Power supplied = Power Consned Energy is conserved, Ps - V.I = 46xts) - ~250w Po = 2/5)² = sow Ew = 36+72+ outu8-230+50 Scanned By SelfScan