Assume the resistance values are R1 = 2,500 , R2 = 1,200 , R3 = 4,600 , and R4 =
ID: 777115 • Letter: A
Question
Assume the resistance values are R1 = 2,500 , R2 = 1,200 , R3 = 4,600 , and R4 = 5,900 , and the battery emfs are 1 = 1.5 Vand & = 3,0 . Use Kirchhoff's rules to analyze the circuit in the figure below. R1 R2 F1 R3 13 (a) Let 11 be the branch current though R1, I2 be the branch current through R2, and I3 be the branch current through R3. Write Kirchhoffs 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 Kirchhoffs 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 currents. (Use any variable or symbol stated above as necessary.) (d) You should now have three equations and three unknowns (11, 12, and 13). Solve for the three branch currents 12 =Explanation / Answer
Given R1 =2500 ohm ,R2 = 1200 ohm, R3 = 4600 ohm, e1 = 1.5V, e2 = 3 V.
Frm Kirchhoff's voltage rule,in any closed loop the algebraic sum of the emf and the potential drops across the resistors and some other elements is zero
and we can take the potential drop across the resistor is -ve if we are in the same direction as that of current in the loop.
a)
Applying kirchhoff's voltage law for left side loop
e1 -i1*R1-i2*R3= 0 ---------(A)
b)
for hte right loop
e2-i2*R2-i3*R3 = 0 ---------(B)
the current loop at the junction A the sum of the currents at the junction is zero or the total currents entering the juction is equal to the sum of the currents leaving the junciton .
c)
i1+i2 = i3 --------------(C)
now substituting the values in the above equations
(A) ===> 1.5-i1*2500 -i3*4600 = 0
(B) ===> 3-i2*1200-i3*4600 = 0
(C) ==> i1+i2 = i3
d)
solving the equations for i1,i2,i3 we get
i1 = -0.2547 mA
i2 = 0.7193 mA
i3 = 0.4645 mA