(a) What is the equivalent resistance of the circuit shown in Fig. 19-70 ? (b) What is the curre

step 1 This is the diagram given in the problem. To make work more simple, we will assign different nam

(a) What is the equivalent resistance of the circuit shown...

Key Concepts

Series and Parallel Resistors

This concept involves identifying and combining resistors that are connected either in series, where resistances add directly, or in parallel, where the reciprocal of the total resistance is the sum of the reciprocals of the individual resistors. Mastering these combinations is essential for simplifying complex circuits into an equivalent resistor that represents the overall resistance of the network.

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits which states that the voltage across a resistor is directly proportional to the current flowing through it, with the resistance being the constant of proportionality (V = IR). This relationship is crucial for finding unknown quantities in circuits, such as current through a resistor, given the voltage drop and resistance.

Equivalent Resistance

Equivalent resistance refers to the single resistance that can replace a more complex network of resistors without changing the overall current or voltage characteristics of the circuit. This simplification makes analyzing the circuit easier, particularly when determining total current, voltage distribution, or power consumption.

Current Division

The current division rule is used to determine the distribution of current across parallel branches in a circuit. According to this rule, the current through a particular branch is inversely proportional to the resistance of that branch compared to the total parallel resistance, which is vital for analyzing circuits with multiple parallel pathways.

Power Dissipation in Resistors

Power dissipation in a resistor is the conversion of electrical energy into heat, which can be calculated using formulas such as P = I²R or P = V²/R. Understanding how to calculate power dissipation is important for ensuring that components can handle the heat generated and operate safely within their specified limits.

Transcript

00:02 This is the diagram given in the problem.

00:07 To make work more simple, we will assign different names to registers.

00:16 Let 12 -o -home register is named as r1, 30 -home register is named as r2, 4 .5 -home register as r3, and 18 -home register as r4.

00:34 And in part a of this problem we want to find equivalent resistance of the circuit here resistance r1 and r2 are in parallel combination so first we will find their net resistance by using the relation r12 equal to 1 upon r1 plus 1 upon r2 hold to the power minus 1 by putting different values here 1 upon r1 is 12 om plus r2 is 30 om this gives us their net resistance equal to 8 .57 om now the diagram will become like this this is battery of sequence volts this is r4 this is r3 and this is the net distance of r1 and r2 and it is r12 again by seeing this diagram it is clear that r12 and r3 are in series so we can find their net resistance by using the relation r1 2 3 equal to r12 plus r3 sr12 is 8 .57 and r3 is 4 .5 om this gives us r123 equal to 13 .07 now that simple form of this diagram will become this is a battery of 6 volt this is r4 and this is r1 2 3 here r1 2 3 and r4 are in parallel combination so in parallel combination we can write equivalent resistance that is a reciprocal of equivalent resistance equal to 1 upon r1 2 3 plus 1 upon r4 this gives us r equivalent equal to r123 into r4 divided by r123 plus r4 now by putting different values here as r123 is 13 .07 om into r4 is 18 om divided by 13 .07 om plus 18 om this gives us equivalent resistance of circuit equal to 7 .57 now in the second part of this problem we want to find the amount of current passing through 18 -on register here 18 -on register is connected across the terminals of battery so voltage drop across 18 -on register is equal to voltage of battery so we can write voltage drop across 18 -home register is equal to current through 18 -oam register into its resistance this gives us current through 18 -oom register equal to voltage drop across 18 -oom register divided by r4 s voltage across 18 -oom register is equal to voltage of battery that is 6 .0 volt divided by resistance of r4 is 18 -oom this gives us current flowing through 18 -home register equal to 0 .33am here we will again consider this diagram here r1 2 is the net resistance of r1 and r2 r1 2 r1 2 and r2 and r3 are in series connection to make the diagram more simple we will assign different names to points let this is named as point a this is named as point b this is c this is now here we will take the loop d a b, c, d, then back to a here we will write loop equation starting from point d and moving counterclockwise so we can write 6 .0 volt minus a motor of current passes through r1 2 let this is i r1 2 i r1 2 into i r1 2 into resistance is 12 om minus s r12 and r3 are in series so current passes through r3 will also be equal to i r12 into r3 is 4 .5 ome equal to 0 this equation gives us the current through resistance r1 2 and r3 equal to 0 .459 ampere so we can say that amount of current passes through r3 that is a motor of current passes through 4 .5 home register equal to 0 .459 ampere...

Please give Ace some feedback