Question

Refer to the figure. 1 = ? Vi = ? R = 1.9012 V = 12.0 V b = ? R = 6.40 R = 14.0 22 • V = ? (a) Calculate P3 (in W). W (b) Find the total power in W) supplied by the source. w Compare the total power with the sum of the powers dissipated by the resistors. (Consider two values to be equal if they are within 3% of each other.) The total power is equal to the sum of the powers dissipated by the resistors. The total power is less than the sum of the powers dissipated by the resistors. The total power is greater than the sum of the powers dissipated by the resistors.

Answer 1

Solution - Hou I CL Y = ? MW R = 1.gor Now equivalent resistance- VE12.0V La R=Ri+ Ryx Rs Rzt R2 P=6.42} R3 -14.0r V? = 1.9 +1 6.4 x 14 6.4+24 otrol = 1.9 +4.392 Daun (R= 6-292 2 Dort 21 bilgqua So- circuit the current (I) flowing in the (I) = ser - 12.0 6.2920 - 191 sat oh bougs & Istot AV botegi I = 1.91 A 2 rotingut So- voltage dissipated in R =) V = IR SA = 1.91 X 109 So v=3.623 volt the voltage across R2 & R3 will be- v's 12.0-3.623 v'- 8-376 volt " R&R3 are connected parallel som the voltage in each resistor will be same.

So- V₂ = 8.376 volt prenature Part Cal - Power (P.₂) 200 2 " V3 3 oupa won R3 połu = (8.376)? 9 14 བ་ 2. 5.011 Watt Part(b) in the total power Ain By wice supplied the source P= vI ni Die - 12.0x1.91 IP = 22.92 watt ON Am sum The total power is equal to the of the powers dissipated by the resistors Spotlod beto Am