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enter image description here

I was having an issue with the solution of this problem. I used mesh analysis with the equations, where \$i_{lamp}=i_l\$, \$i_{oven}=i_o\$, \$i_{heater}=i_h\$:

\$-120+120(i_l-i_o)=0 - (1)\$

\$-120+10(i_h-i_0)=0 - (2)\$

\$8i_o+10(i_o-i_h)+120(i_0-i_l)=0 - (3)\$

Solving these equations, I got \$i_o=30A\$, \$i_l=31A\$, \$i_h=42A\$. However, the answers did the following: enter image description here

I do not understand why my analysis does not work. I also do not understand how the solutions method of assuming a 120V potential difference across the lamp and heater resistors is valid.

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  • \$\begingroup\$ In your first mesh equation, you wrote the current across the lamp, I_L, is equal to I_L-I_O. Isn't it a contradictory statement to say that the current across the lamp is equal to the current across the lamp minus the current across the oven? Same for the other mesh equations as well. \$\endgroup\$
    – xrosaber
    Commented Mar 11, 2023 at 6:38
  • \$\begingroup\$ Sorry, I am very confused, my first equation was -120+120(i_l-i_0)=0. How does that state i_l=i_l-i_o? \$\endgroup\$
    – Jason2134
    Commented Mar 11, 2023 at 6:44
  • \$\begingroup\$ @Janson2134 Mesh analysis is based on KVL; the sum of voltages in a loop is zero. So in your first mesh equation, you are summing the voltage "drops and rises" in the loop with the battery and lamp. Going counterclockwise in the loop like you have, you get a "voltage drop" going from the positive terminal of the battery to the negative terminal of the battery (falling down), hence -120V. \$\endgroup\$
    – xrosaber
    Commented Mar 11, 2023 at 7:10
  • \$\begingroup\$ Since you've assumed the current flowing through the lamp to be from top to bottom, the +ve terminal is on top of the lamp and -ve terminal on the bottom of the lamp (since conventional current goes from higher potential to lower potential). Now the voltage across the lamp can be calculated by V_L=I_L x R_L. Again, going counterclockwise in the loop like you have, you get a "voltage rise" going from the -ve terminal of the resistor to the +ve terminal of the resistor (climbing up) , hence +V_L. \$\endgroup\$
    – xrosaber
    Commented Mar 11, 2023 at 7:10
  • \$\begingroup\$ Now summing up the voltages in loop with the battery and lamp, you get -120 + V_L= -120 + (I_L x R_L)=0. Since, the sum of voltages in a loop is zero. Now notice that writing V_L as R_L(I_L - I_O) implies that the current through the lamp is the difference of the current through the lamp and the current through the oven. \$\endgroup\$
    – xrosaber
    Commented Mar 11, 2023 at 7:10

1 Answer 1

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I've just added your loops, below.

enter image description here

Please do take note of something, though. \$I_{_\text{HEATER}}\ne I_{_\text{H}}\$ and \$I_{_\text{LAMP}}\ne I_{_\text{L}}\$. (It is true that \$I_{_\text{OVEN}}= I_{_\text{O}}\$, though.)

Instead, \$I_{_\text{HEATER}}= I_{_\text{H}}-I_{_\text{O}}\$ and \$I_{_\text{LAMP}}= I_{_\text{L}}-I_{_\text{O}}\$.

Your equations came up with the right results. \$I_{_\text{O}}=30\:\text{A}\$, \$I_{_\text{H}}=42\:\text{A}\$ and \$I_{_\text{L}}=31\:\text{A}\$. Nothing wrong there. You just then needed to realize (which you did when setting up the mesh in the first place) that the net current through the heater and the lamp is the difference between those loop currents.

So \$I_{_\text{HEATER}}= I_{_\text{H}}-I_{_\text{O}}=42\:\text{A}-30\:\text{A}=12\:\text{A}\$ and \$I_{_\text{LAMP}}= I_{_\text{L}}-I_{_\text{O}}=31\:\text{A}-30\:\text{A}=1\:\text{A}\$.

The mesh loop currents need to be combined when a device carries more than one loop current.

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