Use the circuit diagram as Fig. 2 of Assignment 7. R, = 420 Ohm, R_2 - 870 Ohm a
ID: 1456036 • Letter: U
Question
Use the circuit diagram as Fig. 2 of Assignment 7. R, = 420 Ohm, R_2 - 870 Ohm and R_3 = 380 Ohm, and R_4 = 510 Ohm. epsilon_1 = 4.50 V, epsilon_2 = 9.00 V, and epsilon_3,=6.0 V. Determine the current in R_2. Answer: 2.97 mA. 2.44 A. 6.54 mA. 1.97 mA. 3.87 mA. Determine the power dissipated in R_2 in the problem. Answer: 0.037 W. 1.77W. 6.78W. 8.97 J. 1.37 J. Determine the current in R_3. Answer: 2.73 Times 10^(-1) A. 1.77 Times 10^(3) C. 6.78 Times 10^(-3) C. 5.43 Times 10^(-3) A. 1.37 Times 10^)-2) A. Determine the potential difference between c and b in the figure. Answer: 4.73 V. 1.77 V. 6.78 V. 8.97 V. 2.44 V. Determine the power dissipated in R_4. Answer: 1.77 Times 10^(-3) W. 0.625 Times 10^(-3) W. 6.78 Times 10^(-1) J. 8.97 Times 10^(2)W 1.37 Times 10^(-2) JExplanation / Answer
let the current through E3 or current in R3 = I and direction is left to right
and current in R1 and R2 = I1 and direction is left to right
so current is R4 = (I - I1)
now kirchhoff law in upper loop
I1*R1 + I1*R2 - R4*(I - I1) - E2 = 0
1290*I1 - 510*I + 510*I1 - 9 = 0
1800*I1 - 510*I - 9 = 0 ... (1)
kirchhoff law in lower loop
- E3 + E2 + R4*(I - I1) + R3*I - E1 = 0
- 6 + 9 + 510*I - 510*I1 + 380*I - 4.5 = 0
890*I - 510*I1 - 1.5 = 0
multiply this eq by (510/890)
510*I - 292.25*I1 - 0.86 = 0 ... (2)
now add eq(1) and eq(2)
I1*(1800 - 292.25) - 0.86 - 9 = 0
I1 = 0.00654 A
put this value in eq (2)
510*I - 292.25*0.00654 - 0.86 = 0
I = 0.005434 amp
current in R2 = 0.00654= 6.54 mA
power dissipated in R2 = I^2*R = (0.00654)^2*870 = 0.037 watt
current in R3 = 0.00543 = 5.43 mA
potential difference between c and b = Va - Vb = Vab
now using kirchhoff law
Va - E3 + E2 + R4*(I - I1) = Vb
Va - Vb = E3 - E2 - R4*(I - I1) = 6 - 9 - 510*(0.005434 - 0.00654) = - 2.44 volt
current in R4 = I1 - I = 0.00654 - 0.005434 = 0.001106 amp
power = (0.001106)^2*510 = 0.000624 = 0.624*10^(-3) W
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