For the circuit in problem 11 (Figure 5-14), if the voltage across the parallel branches is chan

For the circuit in problem 11 (Figure 5-14), if the voltage...

Key Concepts

Ohm's Law

Ohm's Law is the fundamental principle that describes the relationship between voltage, current, and resistance in an electrical circuit (I = V/R). It is essential for understanding how changing the voltage across a circuit will proportionally affect the current, assuming the resistance remains constant.

Equivalent Resistance in Parallel Circuits

When resistors are connected in parallel, the overall or equivalent resistance of the circuit is determined by taking the reciprocal of the sum of the reciprocals of the individual resistances. This concept is crucial because it simplifies complex circuits into a single resistance value, making it easier to analyze how changes in applied voltage influence the total current.

Kirchhoff's Current Law (KCL)

Kirchhoff's Current Law states that the total current entering a junction in a circuit must equal the total current leaving the junction. In parallel circuits, this principle is used to understand how the current splits among different branches, leading to the total current being the sum of the currents through each branch.

Impact of Voltage Scaling on Circuit Performance

When the voltage applied to a circuit is altered, the resulting currents in the circuit change in accordance with the circuit's resistances and configuration. Doubling the voltage, for instance, will generally double the current through each branch in a circuit with constant resistance, assuming linear behavior as described by Ohm’s Law.

Transcript

00:02 Hello, so in this problem i have my circuit on the right labeled mnopqr i have my currents i 1 i2 i3 so to find the individual current i'm going to use catch -off's law which is at the sum of the emfs in the loop because the potential difference so for the first part i took the loop mnopqm that the bigger loop and the so in that loop the total emf is 10 volts and the individual potential differences is given as that so i have i1 going through the 2 oom and the 3 home and i have i 3 3 going through the 10 oom resistor and i'm going on to clockwise so i'm going to get that equation which is where is that so that's equation one then i go to another loop r -o -p -q -r i go and clockwise again and that gives me equation two at the junction that's at r as a junction rule you can see that i -1 plus i -2 gives you i -3 so i have that equation as well so that's equation 3 so this so i made i isolated i 2 to give me i 3 minus i 1 and i put that in equation 2 so i place that here so whenever i see i 2 i'm going to put i 3 minus i 1 and that gives me that so that gives me this equation right here so that's the 20 that which is also the same as this one the same thing then i bring back equation one which is 10 volts because that so now i have two equations i can solve simultaneously because i have five i and i have five i one is positive one is negative so i just added the two equations so the left side gives me 20 plus 30 then that's 30 the eyes will cancel out and i have 25 i3 so i get i3 to be 1 .2 i go back to use my previous equations to find i1 i get negative 0 .4 as just showing the direction in negative and i get i2 to be 1 .6 so these are the individual current in the branches if we go to the second part which is to find the potential and ab so ab in the diagram so that is a and that's b right there so if you want to find the potential v we want the current going through the 3 om and going through the 4 om so for the 3 om we have i 1 and for the 4 om we have i 2 and so if we go to b so that's just the potential across the 3 oom plus the potential across the 4 om.

04:29 And that gives you 7 .6 volts...

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