Three resistors are joined together across a 30 Volts battery (see the figure).

Question

Three resistors are joined together across a 30 Volts battery (see the figure). The voltage drop across resistor R1 is 8.395 Volts, the current i2 through resistor R2 is 0.312 A, and the power dissipated in resistor R3 is 4.605 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

VR1=8.395 V, IR2=0.312 A, WR3=4.605 W, resistors in series to the

voltage source(I assume), Vbatt=30 V

a) R1=8.395 V/0.312 A26.91

R2=30 V-(VR1+VR3)=30 V-(8.395+14.76)6.845 V/0.312 A21.94

R3=P/I2=4.605 W/0.3122 A47.31

b) Ieq=if all three are in series, than they all have 0.312 A through them

c) PR1=E*I=8.395 V*0.312 A2.62 W

PR2=6.845 V*0.312 A2.14 W

d) Req=for a series circuit:(26.91+21.94+47.31) 96.153 or 96.2

e) 0.312 A

f) Weq=(2.62+2.14+4.605) W9.36 W

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