Question

Using Kirchhoff's junction rule, determine which one of the following four diagrams would be the appropriate one to use for solving this problem. Calculate the current I2 flowing in emf source E2.

Part A

Using Kirchhoff's junction rule, determine which one of the following four diagrams would be the appropriate one to use for solving this problem.

Part B

Calculate the current I2 flowing in emf source E2.

Answer 1

According to Kirchhoff's junction rule, the sum of currents towards the junction is equal to the sum of currents away from the junction. According to Kirchhoff s loop rule, the algebraic sum of the potentials across all the components in a closed loop is zero. Figure A: The current I, is coming from battery a, and the current I, coming from the battery £ According to junction rule, the current (1, -/,) should be passed through the resistor R Thus, the figure A is not correct. Figure B: According to junction rule the current I, should passed through the resistor R. Thus, the figure B is not correct Figure C: According to junction rule the current (1, -/,) should be passed through the resistor R, , and the current I is passed through resistor R. Thus, the figure C is not correct Figure D: According to junction rule the current (1, -/,) should be passed through the resistor R, , and the current I, is passed through resistor R. Thus, the figures corect

Apply Kirchhoff's loop rule for the top loop in the figure D, Apply Kirchhoff's loop rule for the bottom loop in the figure D Rearrange the equation El-IM-1272-IR-0 for 11 1 1'1 22 1 Substitute this value in equation -6, +(1-I)R, to obtain the expression for I,, 8, -1 2-1 R,-

Answer 2

Solution

Part a)

For Figure A, applying Kirchhoff's junction rule we have that:

$I_{2}+I_{2}=I_{1}-I_{2}\, \, \, \, \, \, \, \, \Rightarrow \, \, \, \, \, \, \, \, \, \, I_{2}=I_{1}$

Therefore, figure A is not correct.

For figure B, applying Kirchhoff's junction rule we have that:

$I_{1}+I_{2}=I_{2}-I_{1}+I_{2}\, \, \, \, \, \, \, \, \Rightarrow \, \, \, \, \, \, \, \, \, \, I_{1}=I_{2}-I_{1}$

therefore, figure B is not correct

For figure C, applying Kirchhoff's junction rule we have that:

$I_{2}=I_{2}+I_{2}-I_{1}\, \, \, \, \, \, \, \, \Rightarrow \, \, \, \, \, \, \, \, \, \, I_{1}=I_{2}$

therefore, figure C is not correct.

For figure D, applying Kirchhoff's junction rule we have that:

$I_{1}-I_{2}+I_{2}=I_{1}\, \, \, \, \, \, \, \, \Rightarrow \, \, \, \, \, \, \, \, \, \, I_{1}=I_{1}$

Therefore, the correct figure is figure D:

2 2

Part b)

と 2 2 2

Applying Kirchhoff's loop rule at loop 1 we have:

$-V_{1}+I_{1}R+V_{2}=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (1)$

but:

$V_{1}=\varepsilon _{1}-I_{1}r_{1}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (2)$

and

$V_{2}=\varepsilon _{2}-I_{2}r_{2}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (3)$

Replacing (2) and (3) into equation (1):

$-\varepsilon _{1}+I_{1}r_{1}+I_{1}R+\varepsilon _{2}-I_{2}r_{2}=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (4)$

or

$I_{1}\left ( R+r_{1} \right )-I_{2}r_{2}=\varepsilon _{1}-\varepsilon _{2}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (5)$

Simillarly, applying Kirchhoff's loop rule at loop 2 we have:

$-V_{2}+(I_{1}+I_{2})R_{b}=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (6)$

Replacing (3) into equation (6):

$I_{1}R_{b}+I_{2}(R_{b}+r_{2})=\varepsilon _{2}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (7)$

Grouping equations (5) and (7) we have:

$\left\{\begin{matrix} I_{1}\left ( R+r_{1} \right )-I_{2}r_{2}=\varepsilon _{1}-\varepsilon _{2} & & & \\ I_{1}R_{b}+I_{2}(R_{b}+r_{2})=\varepsilon _{2}& & & \end{matrix}\right.$

Solving the system for $I_{2}$ we have:

${\color{Blue} I_{2}=\frac{\varepsilon _{2}(1+R_{b})-\varepsilon _{1}R_{b}}{r_{2}R_{B}+\left ( R_{b}+r_{2} \right )\left ( R+r_{1} \right )}}$

Answer 3

Part-A The junction rule states that the current coming towards the junction is always equal to the current going away from the junction Figure-A The junction at negative side of emf e, does not satisfy the junction rule. Since current going towards the junction is 21 while, current going away from the junction is (-1,) Therefore, the junction rule is not satisfied. Figure-B The junction at negative side of emf E, does not satisfy the junction rule. Since current going towards the junction is (-I-1) while, current going away from the junction is I. Therefore, the junction rule is not satisfied. Figure-C The junction at negative side of emf e, does not satisfy the junction rule. Since current going towards the junction is I while, current going away from the junction is (21,-1) . Therefore, the junction rule is not satisfied. Figure-C The junction at negative side of emf e, does not satisfy the junction rule. Since current going towards the junction is (I, +I-1, 1)while, current going away from the junction is I. Therefore, the junction rule is satisfied. Hence, circuit in figure-A is appropriate to use.

Part-B Use the Kirchhoff's loop law in upper loop of the circuit in figure-A -E2-1R-Ii-I,r 0 Rearrange the above-mentioned equation for I (1) -(-) (R+ Use the Kirchhoffs loop law in lower loop of the circuit in figure-A. -(1-1)R, +I,y =0 (2) Substitute value of I, in equation-2 (4-)-1, (R+) -I R, +I0 Rearrange the above-mentioned equation for I F-(3-) (R, ) (R+)(R)