Four resistors, R_{1} = 18.1 Ω, R_{2} = 37.8 Ω, R_{3} = 91.6 Ω and R_{4} = 19.0 Ω are connected to a 18.0 V battery as shown in the figure. Determine the power dissipated by resistor R_{2}.
1/R123 = 1/(R1+R2) + 1/R3 = 1/55.9 + 1/91.6 = 0.028806
R123 = 34.71 ohm
and
Rt = R123 + R4 = 34.71 + 19 = 53.71 ohm
so total current is
I = V/R = 18.0/53.71 = 0.3351 A
and PD across R4 is
V = I R4 = 0.3351 (19) = 6.37 v
so PD across R1_R2 branch is
12.0 – 6.37 = 5.63 v
and current through that branch
I = V/R = 5.63 / 55.9 = 0.1007 A
finally
P = I^2 R = 0.1007^2 (37.8) = 0.384 watts
Kindly rate:
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