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 7.471 Volts, the current i2 through resistor R2 is 0.199 A, and the power dissipated in resistor R3 is 2.736 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?

Find the total resistance of the network.

Find the current drawn from the battery.

Find the total power used by the circuit.

Explanation / Answer

you have R2 & R3 in parrallel and then R1 in Series To add the 2 parrallel resistors (you need to do this first) 1/Rparellel = 1/R2 + 1/R3 Rparellel = 1/ (1/R2 + 1/R3) Then for the series bit you just add them together So Rtotal = R1 + Rparellel so V2=V3=VT-V1=24-7.471=16.529 V P=VI so P3=V3*I3 I3=P3/V3=2.736/16.529=0.1655 A then R3=V3/I3=16.529/0.1655=99.87 ohms we know the value V2 then R2=V2/I2=16.529/0.199=83 ohms P2=V2*I2=16.529*0.199=3.289 W I1=I2+I3=0.199+0.1655=0.3645 A R1=V1/I1=7.471/0.3645=20.5 ohms power dissipated across R1= P1=V1*R1=7.471*20.5=153.15 W The total power PT=P1+P2+P3=153.15+3.289+2.736=159.17 W The total resistance is RT=VT/I=24/0.3645=65.8436 ohms current drawn from the battery=VT/RT=24/65.8436=0.3645 A :)

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