A battery with E=6.00 V and no internal resistance supplies...

A battery with $\boldsymbol{E}=6.00 \mathrm{V}$ and no internal resistance supplies current to the supplies current to the circuit shown in Figure P 18.14. When the double-throw switch $S$ is open as shown in the figure, the current in the battery is 1.00 $\mathrm{mA}$ . When the switch is closed in position $a$ , the current in the battery is 1.20 $\mathrm{mA}$ . When the switch is closed in position $b,$ the current in the battery is 2.00 $\mathrm{mA}$ . Find the resistances (a) $R_{1},$ (b) $R_{2},$ and ( $) R_{3}$ .

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits that states the relationship between voltage, current, and resistance. It is commonly expressed as V = IR, where V is voltage, I is current, and R is resistance. This law enables one to calculate any one of the three quantities provided the other two are known, and it serves as the basis for analyzing both simple and complex circuits.

Series Circuits

In a series circuit, electrical components are connected sequentially, so the current that flows through one component must flow through all others. The total resistance in a series circuit is the sum of all individual resistances. This concept is critical for analyzing circuits where resistors are arranged end-to-end, as it simplifies the calculation of the overall voltage drop and current flow.

Parallel Circuits

In a parallel circuit, components are connected across the same two points, providing multiple paths for the current to flow. The voltage is the same across each component, while the total current is the sum of the currents through each individual path. Understanding parallel circuits is essential for breaking down complex networks and determining how resistors affect the overall behavior of a circuit.

Equivalent Resistance

Equivalent resistance is a concept used to simplify complex resistor networks by finding a single resistor that can replace a group of resistors without changing the total current drawn from the voltage source. In series circuits, the equivalent resistance is the sum of resistances, while in parallel circuits, reciprocal calculations are used. This simplification is key to applying Ohm's law and Kirchhoff's circuit rules in more intricate scenarios.

Circuit Analysis Techniques

Circuit analysis techniques, including the use of Kirchhoff’s Voltage and Current Laws, allow for a systematic approach to solving complex circuits. These laws help in writing equations based on the conservation of energy and charge around any closed loop or junction in the circuit. These techniques, combined with the concepts of Ohm’s Law and equivalent resistance, enable one to determine unknown circuit parameters and validate the behavior of electrical networks.

Transcript

00:01 In this particular circuit we need to find the values of r1, r2 and r3 registers.

00:07 Now, so step a in which as for the given condition, value of the current which is flowing through the circuit, that is i, is equal to 1 mm.

00:22 Now when the switch is open, the current through vertical resistor r2 is 0.

00:29 So this particular resistor r2, which is connected towards a and the other, node is 0.

00:36 Why? because it is open -ended, so current will not flow through it.

00:41 So we'll first try to draw the diagram to show the connections.

00:48 So here first, epsilon, which is the emf, which is of 6 volt.

00:59 One resistor which is r1 in series with r2, the vertical resistance r2 should not be taken into consideration now because the current will not be flowing through it since it is open -ended so r1 r2 and r3 will be in series and the current which is flowing through them is of 1 millie -amper as per the given condition so simply by using oms law epsilon is equal to i into combination of resistances which is r1 plus r2 plus r3 because they are connected in series so so therefore 6 volt is equal to 1 millie ampere in multiplication with the sum of 3 resistances.

02:03 Therefore, r1 plus r2 plus r3 is equal to 6 volt divided by 1 millie ampere which is equal to 6 ,000 and we'll consider this as equation 1.

02:33 We'll go for the next step.

02:37 Step b.

02:40 The condition is when switch is at position a, the amount of current which is flowing to the circuit is of 1 .2 millie -amperors.

02:59 So we'll draw the circuit first.

03:04 This is of 6 volt.

03:07 R1 will be there in series of 6.

03:13 With r2.

03:14 Since the switch is at position a, r2 vertical will be into consideration and our circuit will be like this.

03:29 So the vertical r2 is connected between c, a and d points.

03:37 Rather we consider them as nodes.

03:40 Now in this case, both the r2s are connected between node c and node d.