Question

7. (20 points) For the circuit below, find the power dissipated by resistor $R_3$ $R_1 = 2 \text{ k}\Omega$ $R_2 = 330 \Omega$ 45 V $R_3 = 720 \Omega$ $R_4 = 680 \Omega$ $R_5 = 220 \Omega$ 1 cal = 4.186 Joules $C_{water} = 1 \text{ cal/gm°C}$ $C_{ice} = 0.0502 \text{ cal/gm°C}$ $L_f = 79.9 \text{ cal/gm}$ $L_v = 540 \text{ cal/gm}$

          7. (20 points) For the circuit below, find the power dissipated by resistor $R_3$
$R_1 = 2 \text{ k}\Omega$
$R_2 = 330 \Omega$
45 V
$R_3 = 720 \Omega$
$R_4 = 680 \Omega$
$R_5 = 220 \Omega$
1 cal = 4.186 Joules
$C_{water} = 1 \text{ cal/gm°C}$ 
$C_{ice} = 0.0502 \text{ cal/gm°C}$ 
$L_f = 79.9 \text{ cal/gm}$ 
$L_v = 540 \text{ cal/gm}$

7. (20 points) For the circuit below, find the power dissipated by resistor R3
R1 = 2 kΩ
R2 = 330 Ω
45 V
R3 = 720 Ω
R4 = 680 Ω
R5 = 220 Ω
1 cal = 4.186 Joules
Cwater = 1 cal/gm°C
Cice = 0.0502 cal/gm°C
Lf = 79.9 cal/gm
Lv = 540 cal/gm

Added by Sandra C.

Physics

Robert Coleman Richardson; Betty… 5th Edition

Chapter 18

Instant Answer

Solved by Expert Sikandar Baig

Step 1

To find the current flowing through the circuit, we can use Ohm's Law: V = IR, where V is the voltage and R is the resistance. In this case, the voltage is 220V and the resistance is the sum of R1, R, and R4: R_total = R1 + R + R4 = 2kΩ + 330Ω + 680Ω = Show more…

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For the circuit below, find the power dissipated by resistor R. R1 = 2kÎ© R = 330Î© R4 = 680Î© 220V 1 cal = 4.186 Joules Cuter = 1 cal/gmÂ°C cie = 0.0502 cal/gmÂ°C Ly = 79.9 cal/gm L = 540 cal/gm