Question

Five resistors R1=9.00 Ω, R2=7.00 Ω, R3=13.0 Ω, R4=10.0 Ω, and R5=7.00 Ω are mounted as shown in the figure. The voltage of the battery is V = 22 Volts. What is the power dissipated in the R3 resistor?

WWWWW R4 R5 R R3

Answer 1

In order to calculate the power dissipation across resistance R3, we have to calculate the equivalent resistance of the circuit and then find the current through R3.So the power dissipation will be then P3 = I^2R3

Here we were the which is in paraller series combination of fo of Rs connection with the series combination e of we consider the resistence in parallel combination be to, Roa (Pg 1P3) I (PARS) (7.80419,6) (10.047.00) - 20.00 17.000 - 120.0x17.0 39 Again R is in stories with Rp. co equivalent resistance in circult, Ra Art Rp a 9.000 & abon - 18.19 r. so, the current through the circur, 1. * = 22 x = 1.91 À so, the voltage acros b, ya 292 (1:21 AX 9 h) 2 10.89 volts. so woltuge across parallel corredor, apo vrv, a (22-10.89) & lluv. Now if I be the current through of and as and If Iz te the current through ad Ry where [2242 other 2 NP 22 Futrza 7.00 - 13.0) 2 ² a I'll & 0.556 A. This current flows through f & Rz. 40, the power dissipated in as, Pza Ly Rz 10.5564) x 13-0 24.02. W the lower dissipated in the fa ressor is 4.02w.