WA Hw 6 C fi t/web/Student/Assig 13102429 Need Help L Read t Talk to a Tutor 9.

Question

WA Hw 6 C fi t/web/Student/Assig 13102429 Need Help L Read t Talk to a Tutor 9. -6 points GioCP2 19. P.056 My Notes Assume the resistance values are R1 2,600 n, R2 1,600 n, R3 4,400 n, and R4 6,200 n, and the battery emfs are E1 1.5 v and E2 3.0 V Use Kirchhoff's rules to analyze the circuit in the figure below. (a) Let T1 be the branch current though R1, 12 be the branch current through R2, and 13 be the branch current through R3. Write Kirchhoff's loop rule relation for a loop that travels through battery 1, resistor 1, and resistor 3. (Use a clockwise current loop when entering your answer. Use any variable or symbol stated above as necessary.) (b) Write Kirchhoff's loop rule relation for a loop that travels through battery 2, resistor 2, and resistor 3. (Use a clockwise current loop when entering your answer. Use any variable or symbol stated above as necessary.) (c) Apply Kirchhoff's junction rule to the junction at A to get a relation between the three branch any variable or symbol stated above as currents. Use necessary.) (d) You should now have three equations and three unknowns (11, 12, and 13). Solve for the three branch currents. Need Help Read it Talk to a Tutor Submit Answer Save Progress Practice Another Version 10. -12 points GioCP2 19-P.061 My Notes below. What is the current through the circuit just after the switch is closed and after the switch has been closed for a very long t Consider the RC circuit 9:15 PM Ask me anything 8/2016

Explanation / Answer

here,

R1 = 2600 Ohms
R2 = 1600 Ohms
R3 = 4400 Ohms
R4 = 6200 Ohms
Emf, E1 = 1.5 V
E2 = 3 V

Part A:
Let I1 be the branch current though R1, I2 be the branch current through R2, and I3 be the branch current through R3. Write Kirchhoff's loop rule relation for a loop that travels through battery 1, resistor 1, and resistor 3. so

E1 I1*R1 I3*R3 = 0 ---------------------(1)

Part B:
Using Kirchhoff's loop rule relation for a loop that travels through battery 2, resistor 2, and resistor 3.

2 + I3*R3 I2*R2 = 0 ------------------------(2)

Part C:
Appllying Kirchhoff's junction rule to the junction at A to get a relation between the three branch currents.

I1 - I2 -I3 = 0 ------------------------(3)

Using 3 in 1, we get,
(I2+I3)*r1 + I3*R3 = E1

I2*R1 + I3 (R1 + R3) = E1
I2*2600 + I3 (2600 + 4400) = 1.5
I2*2600 + I3*7000 = 1.5 ----------(4)

also Second Eqn Cna be written as,

I2*1600 - I3*4400 = -3 --------------(5)

Comparing 4 and 5, we have two varibale solving for I2 , I3,

Using 5 as I2 = ((-3 + I3*4400)/1600) in 4 we get,

((-3 + I3*4400)/1600)*2600 + I3*7000 = 1.5
upon solvign we get,

I3 = 0.0004505 A -------------------------------------(6)

Using I3 in eqn 4 to get value of I2,

I2*1600 - 0.0004505*4400 = -3
I2 = -0.000636125 A --------------------------------------(7)

Therefore From Eqn 3,

I1 = I2 + I3
I1 = -0.000636125 + 0.0004505
I1 = -0.000185625 A -------------------------------------------(8)

Please check your value of Resistances given in question( They are really high ) , they seems wrong to me,

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.