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

Question

1) Three resistors are joined together across a 24 Volts battery (see the figure). The voltage drop across resistor R1 is 8.401 Volts, the current i2 through resistor R2 is 0.223 A, and the power dissipated in resistor R3 is 2.357 W. What is the value of each resistor? R1?

2) R2?

3) R3?

4) What is the current through each resistor? i1?

5) i3?

6) What is the power dissipated in each resistor? P1?

7) P2?

8) Find the total resistance of the network.

9) Find the current drawn from the battery.

10) Find the total power used by the circuit.

(please show work! thanks!)

Explanation / Answer

v across R1 is 8.401 V==>total V=2424-8.401=15.599 V for the

parallel circuit of R2 and R3==>R3=E2/P=15.5992/2.357103.24

R3 amps=15.599/103.240.1511 A==>R2=15.599/0.22369.951

R2=15.599/69.9510.223 A==>R1=0.223+0.15110.3741 A

R1=8.401/0.374122.46

1) R1=22.46

2) R2=69.951

3) R3=103.24

4) I1=0.3741 A

5) I3=0.1511 A

6) P1=8.401*0.37413.143 W

7) P2=15.599*0.2233.48 W

69.951*103.24/(69.951+103.2441.7

8) R1+parallel combination(41.7)=64.16

9) 0.3741 A

10) P1+P2+P3=3.143+3.48+2.357=8.98 W

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