The circuit shown in Figure P28.21 is connected for 2.00 min. (a) Determine the current in each

The circuit shown in Figure P28.21 is connected for 2.00...

Key Concepts

Kirchhoff's Laws

Kirchhoff's Laws, which include the junction rule (current law) and the loop rule (voltage law), are used to analyze the distribution of current and voltage in electrical circuits. In the context of the circuit problem, these laws help determine the current in each branch and verify that the sum of currents leaving or entering a junction equals zero, and that the sum of potential differences around any closed loop equals zero.

Ohm's Law

Ohm's Law is the fundamental relationship between voltage, current, and resistance in a circuit. It allows one to calculate the current through a resistor given its resistance and the voltage across it. This principle is critical when determining the current in each branch of the circuit as well as the power dissipated by each resistor.

Power and Energy in Electrical Circuits

The relationship between power, voltage, current, and time is essential for understanding how energy is delivered and used in an electrical circuit. Power is defined as the product of voltage and current, and when multiplied by time, it gives the energy delivered. This concept is used to calculate the energy delivered by batteries as well as the energy dissipated by resistors over a given period.

Energy Transformation in Batteries

Batteries transform chemical energy into electrical energy, which then powers the circuit. This energy conversion is a key aspect of circuit operation, where the stored chemical energy in the battery is converted into electrical energy that subsequently does work or is lost as heat in the resistors.

Joule Heating in Resistors

Joule heating refers to the process in which electrical energy is converted into thermal energy (heat) as current flows through a resistor. This phenomenon explains the energy conversion within the resistors, where the electrical energy is transformed into internal energy, typically in the form of heat, signifying energy losses in the circuit.

Transcript

00:01 In this question, we are given this circuit setup and it is connected for two minutes.

00:07 And in part a, you need to determine the current in each branch of the circuit.

00:13 So what i've done is to mark down i1, i2 and i3 in each branch.

00:21 And so i'm going to solve for i1, i2 and i3 using kerchop's rule.

00:30 Okay, so using the junction rule, okay, we have i1 equals to i2 plus i3.

00:42 Okay, so this is our equation one and then i'm going to use loop rule for this for the right loop and also and look two for the outer loop.

00:57 Okay, so using loop rule, okay, loop one.

01:07 We have 12.

01:11 Minus 4 i1 minus 6 i2 minus 4 equals to 0 so we have 8 equals to 4 i1 plus 6 i2 and then simplify further we have 4 i1 plus 3 i2 okay so this is equation 2 and then for loop 2 the equation is the equation is 12 minus 4 i1 minus 8 i3 equals to 0 so we have 3 equals to i 1 plus 2 i 3 okay so this is our equation 3 right so okay so what we can do now is to have to have so we have three equations i1 equals to i2 plus i3, 4 equals to 2 i1 plus 3 i2, and then equation 3 is 3 equals to i 1 plus 2 i 3.

02:38 So what i'm going to do is to sub 1 into 2 and 3.

02:46 Okay.

02:49 Okay, so what i have would be 4 equals to 2 i2 plus i3 plus 3 i2 equals to 5 i2 plus 3 i 2 plus 3 i 2 equals to 5 i 2 plus 2 i3 okay and then i'm also going to have 3 equals to i 1 plus i i 2 plus i 3 plus i 3 which is which we get i 2 plus 3 i 3 okay so i'm going to call this equation 4 equation 5 okay so 5 times 5 minus 4 i'll get 11 equals to the thin i3 this gives i3 equals to 11 over 13 this is 0 .8 4 6 ampiers and then i 2 is equal to 3 minus 3 i 3 okay so this is equal to 0 .462 ampiers and then i1 is equal to i2 plus i3 this is 1 .31 ampere okay so these are the answers for part a okay i3 is 0 .846 ampies i 2 is 0 .46mpies i 2 is 0 .462 ampiers okay in next in part b okay so part b we need to find an energy delivered by each battery okay so the energy delivered by the 12 volt battery is equal to the power times times time okay so the power is i one times e mf of the battery times delta t so i1 is 1 .31 emf is 12 delta t is 2 minutes but in seconds so we need to multiply by 60 so the answer is 1 880 juice okay and then we repeat for the 4 volt battery then energy delivered by the forward battery.

05:52 So this is again, p times delta t.

05:57 But this time you have minus i2e delta t...

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