Question

(a) Kirchhoff's rules
using each circuit in the figure
Find currents passing through the resistance.
(b) between points c and f
by calculating the potential difference (Vc-Vf)
which point is higher
Determine the potential.
(c) Provided by batteries
What is the total power?

I 4.00 k22 b d E E, R3 Ez 70.0 V 60.0 V 80.0 V 12 I 11 R2 3.00 k22 2.00 k92 ww R1

Answer 1

442 M Tov Gov 80V 70-2000 - 60=0 IY 4000 I t 80-3000I -60=0 I, Anf From Kirchoff's law, the ans are In loop 70/2008 I 66=0 4000 I180-30001 3000 I2 -60 = 0 ( Resistances 70- 2000T - 60 =0 40001, +80-3000 I I= It Is

These three equations are to be solved. (o- 27 3 I ( All inke 4 I, + 20-3 I ₂ = 0 Renstances I=It I 10=2LIt I₂) + 3 I = 10 = 2I + 51₂ x 2 20- -4 I, + 3 I ROL 20+20= BIL I= 40 13 3.077AMA Il- 10-5 Iz 10-563.077) 2 -2.69mA. I=0.387 MA Through R t I 3. 077 AMA Through Rs I = -2.69 AMA Through R, = 1 = 0.387 M A Ve +60 + Il tuo = vf V. - Vf - 6o-I₂R2 - (-60 - 3.077 X 3) 69 2310 A VE Vc and potential difference = 69.23) v

O Total powe power by batteries to x I + 60 x + 80XI, 3.077 70x0.387 + 608 269 + 80 x 2.69 426 914x 103 = 0.426 W