Three resistors, R1=10.0 ΩR2=20.0 Ω, and R3=30.0 Ω, are connected in a multiloop circuit, as sho

step 1 Hi, in the given problem there is a combination. A multi -loop circuit. The first loop is having

Three resistors, R1=10.0 ΩR2=20.0 Ω, and R3=30.0 Ω, are...

Key Concepts

Kirchhoff's Circuit Laws

Kirchhoff's Circuit Laws, comprising Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL), are fundamental for analyzing multiloop circuits. KCL states that the sum of currents entering a node equals the sum of currents leaving it, while KVL ensures that the sum of voltage drops around any closed loop is zero. These principles allow for the systematic determination of current and voltage distribution in circuits with multiple loops.

Ohm's Law

Ohm's Law describes the direct relationship between voltage, current, and resistance in an electrical circuit. It states that the voltage across a resistor is equal to the product of the current passing through it and its resistance (V = IR). This law is essential for calculating unknown quantities and forms the basis for analyzing resistive components in any circuit.

Power Dissipation in Resistive Circuits

Power dissipation in a resistor is the rate at which it converts electrical energy into heat. This can be calculated using formulas such as P = I^2R or P = V^2/R, where I is the current through the resistor and V is the voltage across it. Understanding power dissipation is crucial for designing circuits that can safely handle and dissipate heat generated by resistors.

Resistor Networks Analysis

Resistor Networks Analysis involves combining resistors in series and parallel configurations to simplify complex circuits. By finding the equivalent resistance, one can reduce the complexity of the circuit, making it easier to apply Ohm's Law and Kirchhoff's Laws to analyze current distribution and power dissipation.

Transcript

00:01 Hi, in the given problem there is a combination.

00:04 A multi -loop circuit.

00:08 The first loop is having a resistance r1 is equal to 10 .0 om.

00:16 There is a battery providing a potential of v2 is equal to 9 .00 volt.

00:27 In this loop and one more resistor having a resistance of r2 is equal to 20 .0 om.

00:39 Then there is another loop having a resistor providing a resistance of r3 is equal to 30 .0 oom and a battery providing the potential of v1 is equal to 15 .0 volt and this is the complete circuit representing multiple loops now we can name the junctions a b c d e and f now the given currents in the circuit are here this is i1 moving towards right through r1.

01:25 Then there is a current i2 moving again towards right through r2.

01:33 And finally here this is i3 and interestingly that is also moving towards right through r3.

01:41 Now all these three currents are meeting here at the junction e, i1, i2, and i3.

01:55 So first of all, using kirchhoff's junction rule at the junction e we get i1 plus i 2 plus i 3 as all the three currents are entering into the junction but none of the current is leaving so is equal to 0 or we can say equation for i3 is equal to minus i1 plus i 2 we can mark it as equation number 1 now we will apply kirchhoff's voltage law or loop rule so first of all using kirchhoff's loop rule in closed loop a b e f f a, this upper loop, let it be loop first.

03:10 So in this loop if we apply kirchof's loop rule, first of all, it says the algebraic sum of potential drops.

03:19 So the potential will be dropping in this loop at two places.

03:24 First of all, this is i -1r -1 and that is clockwise.

03:29 So it will be taken as positive.

03:31 But for i -2 into r2 that will be taking place counterclockwise.

03:35 So it will be taken as negative and battery v2 is sending its current in counterclockwise direction again so it will also be taken as negative and this kirchop's loop rule says algebraic sum of potential drops means i1 r1 with positive is equal to sorry i 1 r1 positive minus i2 r2 negative is equal to minus v2 now plugging in the known values it becomes 10 i1 the r1 value of r1 is 10 oan 10 i1 minus 20 i 2 is equal to minus 9 so we can mark it as equation number 2 now the same rule kirchop's loop rule in loop closed loop b c d e b b and in this closed loop, the potential drop i3 into r3 counterclockwise.

04:44 The potential drop i2 into r2 here it will be clockwise.

04:49 Battery providing voltage v2 clockwise, battery providing voltage v1, clockwise.

04:56 So using the same kirchops loop here we get i2r2 minus i3r3.

05:09 Is equal to v2 plus v1 or we can say plugging in all the known values this is 20 i2 minus 30 for r3 and i 3 from equation number 1 this is minus i 1 plus i 2 is equal to 15 for v2 v1 is 9.

05:33 So, rearranging the terms we get here, 20 i2 plus 30 i1 plus 30 i2 is equal to 24 or we get it to be 30 i 1 plus 50 i 2 is equal to 24 which is a 20 i 1 plus 50 i 2 is equal to 24 which is equation number 3.

06:10 Now equation number 2 saves 10 i 1 minus 20 i 2 is equal to minus 9...

Please give Ace some feedback