Determine the current in each resistor of the circuit shown in Fig. 19-88. FIGURE 19–88 Problem

step 1 Okay, so this is a very lengthy problem. First, so we have this circuit here and as you can see,

Determine the current in each resistor of the circuit shown...

Key Concepts

Ohm's Law

A fundamental principle that relates voltage, current, and resistance in an electrical circuit using the equation I = V/R. This law is essential for calculating the current through each resistor once the voltage across it is known.

Kirchhoff's Voltage Law (KVL)

A core concept stating that the sum of all voltages around a closed circuit loop is zero. It is used to establish equations that account for voltage drops across components, which is crucial for solving circuits with multiple resistors.

Kirchhoff's Current Law (KCL)

A key principle asserting that the total current entering a junction equals the total current leaving it. This law is important for analyzing node voltages and currents in circuits with multiple branches.

Series and Parallel Resistor Networks

Understanding the configuration of resistors—whether they are arranged in series (same current flows through each) or in parallel (voltage across each resistor is the same)—allows for the simplification of complex circuits into equivalent resistances, making it easier to analyze the current distribution.

Transcript

00:01 Okay, so this is a very lengthy problem.

00:05 First, so we have this circuit here and as you can see, i have marked all the currents that's passing through each registers.

00:15 Now, i mark the current with the value of the resistor corresponding, i mark the current corresponding to the value of the register.

00:26 So for example i6 is the one that's passing through six homes i12 is passing through 12 oms i 6 .8 passing to 6 .8 and so on so forth and now we see we have one two three four five unknowns so we need five equations now we can use junction rule at a and b and then we can use loop rule at left middle and right loop so let's first apply junction rule at point a and b so at point a we see that i top is going away i6 is going away and i5 is going into the junction so that means i 5 should be equal to i6 plus i top similarly in this junction i top is going coming in and i 6 .8 is coming in and then i 12 is going away so that's what we wrote here and if we actually solve for it if we actually substitute itop from both of the equations we get a relation between i56 .8 12 and 6 right so that's that's all about junction rule now for the loop rule what we are doing here is first we're starting with the 17 volts here so these two are in series so that's why we add them and we are taking our direction of current as clockwise and similarly we are taking for the middle one we are taking the direction as clockwise again and for the right loop we're taking the direction as anticlockwise right so here as you can see we're adding 12 and 5 so we we have voltage gain on both batteries so that's why we have 17 volts and there's a plus sign then the current is in the same direction as the loop so that's why you have a negative sign and then we have i 5 times 5 oms then we have negative i 6 times 6 oms because of this current so that's our left loop for the right loop we have two batteries connected in series and then we have a voltage gain so we have 10 volts then i12 times 12 and there's a negative because of the same direction then again 6 .8 is in same direction with the loops direction so it's going to be 6 .8 times i6 .8 and there's a negative sign and for the center loop we see that i6 is in opposite direction with the loops directions or that so sorry it's going to be the opposite but it doesn't matter actually because since there are only two currents will have the same expression but anyway so we see that i s12 is in opposite direction with the loose direction so that's why it's positive and i6 is in the same direction of the loop direction so it's negative but we are not considering this i top current here because uh there's no resistor so that's why there's no voltage drop so we don't have to worry about that all right so now we have technically four equations so we have merged this equation into one so that's why we have four equations let's try to solve them so we start with the fourth equation so you see it's it's the shortest of all so we get a relation between i6 and i12 and then we can use this and substitute i6 on the other equation so that's what we did so when we substituted all the i6 we got expression like this so these are our new equations so these are the three new equations that we have so we got rid of our fourth equation now what we can do here is from there we use equation one or the first one to eliminate i6 .8 so now our first equation is this so we are eliminating 6 .8, yeah, we're eliminating 6 .8 from the first equation.

05:15 So we're writing it in terms of i -12 and i -5.

05:18 So, right, if we substitute back to i -2 and i -3, we get these two following expressions.

05:27 And, okay, so this thing is going to be zero.

05:32 Right.

05:32 So then this is our third equation.

05:34 This is our second equation.

05:36 Now we see that we only have i -5 and i -2...

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