Three resistors are joined together across a 12 Volts battery (see the figure).

Question

Three resistors are joined together across a 12 Volts battery (see the figure). The voltage drop across resistor R1 is 3.429 Volts, the current i2 through resistor R2 is 0.120 A, and the power dissipated in resistor R3 is 0.761 W. What is the value of each resistor? a. R1? b. R2? c. R3? d. What is the current through each resistor? i1? e. i3? f. What is the power dissipated in each resistor? P1? g. P2? h. Find the total resistance of the network. i. Find the current drawn from the battery. j. Find the total power used by the circuit.

Explanation / Answer

I1R1=3.429V......................1
V2/R2=0.120A......................2
V3^2/R3=V2^2/R3=0.761W=I3^2*R3=(I1-0.120)^2*R3.........3
12-I1R1-I3R3=0=12-I1R1-(I1-0.120)R3
I1R1=12-(I1-0.120)R3.......................4
putting 4 in 1
12-(I1-0.120)=3.429
(I1-0.120)R3=8.571V
from 3
0.761=(I1-0.120)*8.571
I1=0.209A
then from 1
0.209*R1=3.429
R1=16.40

from 4

R3=96.30

from 3

V2=8.56V

then from 2

R2=71.34

equivalent resistance of circuit=57.38

power in R2=V2^2/R2=1.027W

power in R1=I1^2*R1=0.1948W

tital power used by the circuit=1.982W

total current drawn from battery=12/57.38=0.2091A

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