Three resistors are joined together across a 24 Volts battery (see the figure).

Question

Three resistors are joined together across a 24 Volts battery (see the figure). The voltage drop across resistor R1 is 8.785 Volts, the current i2 through resistor R2 is 0.223 A, and the power dissipated in resistor R3 is 2.248 W. What is the value of each resistor? R1?

R2?

R3?

What is the current through each resistor? i1?

i3?

What is the power dissipated in each resistor? P1?

P2?

Find the total resistance of the network.

Find the current drawn from the battery.

Find the total power used by the circuit.

 

Explanation / Answer

Given that, The total voltage of the circuit V = 24 V The voltage across resistance R1 is V1 = 8.785 V The current across resistance R2 is I2 = 0.223 A The power dissipated in resistor R3 is P3 = 2.248 W _______________________________________________ The voltage across resistance R1 is 8.785 V and the total voltage of the circuit is 24 V. So the remaining voltage is                       V' = 24 V - 8.785 V                           = 15.215 V From figure, the resistance R2 and R3 are connected in parallel combination. For resistances, in parallel combination the remaining voltage is same for two resistors. that is V' = V2 = V3     ...... (1) Then, power P3 = (V')2/R3 Therefore, the resistance R3 = (V')2/P3                                          = (15.215 V)2/(2.248 W)                                          = 102.98 Hence, the current across resistance R3 is                           I3 = V3/R3                               = (15.215 V)/(102.98 )                               = 0.1477 A ___________________________________________                               = (15.215 V)/(102.98 )                               = 0.1477 A ___________________________________________       Using equation,                      V2 = V3        (since, from Ohm's law V = IR)                    I2R2 = I3R3 Therefore, the resistance R2 = I3R3 /I2                                         = (0.1477 A)(102.98 )/(0.223 A)                                        = 68.20 From figure, the current I1 = I2+I3                                        = 0.223 A + 0.1477 A                                        = 0.3707 A ________________________________________________________ The resistance R1 = V1/I1                          = (8.785 V)/(0.3707 A)                          = 23.7 _________________________________________________________                          = (8.785 V)/(0.3707 A)                          = 23.7 _________________________________________________________ The power dissipated in resistor R2 is                           P2 = I22R2                                = (0.223 A)2(68.20 )                                = 3.39 W ___________________________________________________________ The power dissipated in resistor R1 is                           P1 = I12R1                                = (0.3707 A)2(23.7 )                                = 3.25 W ____________________________________________________________ From figure, the resistance R2 and R3 are connected in parallel combination,                                = (0.223 A)2(68.20 )                                = 3.39 W ___________________________________________________________ The power dissipated in resistor R1 is                           P1 = I12R1                                = (0.3707 A)2(23.7 )                                = 3.25 W ____________________________________________________________ From figure, the resistance R2 and R3 are connected in parallel combination, The power dissipated in resistor R1 is                           P1 = I12R1                                = (0.3707 A)2(23.7 )                                = 3.25 W ____________________________________________________________ From figure, the resistance R2 and R3 are connected in parallel combination,                                = (0.3707 A)2(23.7 )                                = 3.25 W ____________________________________________________________ From figure, the resistance R2 and R3 are connected in parallel combination, so the equivalent resistance is                  R' = R2R3/R2+R3                      = (68.20 )(102.98 )/(68.20 +102.98 )                      = 41.02 From figure, the resistance R1 is connected in series combination with resistance R', so the equivalent resistance R =  R1 + R'                                             = 23.7 + 41.02                                             = 64.72 ______________________________________________________________________ The current draw from the battery is I1 = 0.3707 A _______________________________________________________________________ The total power used by the circuit is         P = P1+P2+P3            = 3.25 W+3.39 W + 2.248 W            = 8.88 W
______________________________________________________ answers:              R1 = 23.7 , R2 = 68.2 and R3 = 102.98              Total resistance R = 64.72              I1 = 0.3707 A, I2 = 0.223 A, and I3 = 0.1477 A            P1 = 3.25 W, P2 = 3.39 W, and P3 = 2.248 W             Total power P = 8.88 W                      = (68.20 )(102.98 )/(68.20 +102.98 )                      = 41.02 From figure, the resistance R1 is connected in series combination with resistance R', so the equivalent resistance R =  R1 + R'                                             = 23.7 + 41.02                                             = 64.72 ______________________________________________________________________ The current draw from the battery is I1 = 0.3707 A _______________________________________________________________________ The total power used by the circuit is         P = P1+P2+P3            = 3.25 W+3.39 W + 2.248 W            = 8.88 W
______________________________________________________ answers:              R1 = 23.7 , R2 = 68.2 and R3 = 102.98              Total resistance R = 64.72              Total resistance R = 64.72              I1 = 0.3707 A, I2 = 0.223 A, and I3 = 0.1477 A            P1 = 3.25 W, P2 = 3.39 W, and P3 = 2.248 W             Total power P = 8.88 W

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.