Three resistors are connected in series across a battery....

Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 2.0$\Omega$ and $4.0 \mathrm{W}, 12.0 \Omega$ and $10.0 \mathrm{W},$ and 3.0$\Omega$ and 5.0 $\mathrm{W}$ . (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in ( a $)$ ?

Key Concepts

Power Dissipation in Resistors

Resistors dissipate electrical energy as heat, and the power dissipated by a resistor can be calculated using formulas such as P = I^2R or P = V^2/R. Understanding this helps in ensuring that no resistor in a circuit is subjected to more power than its maximum rating, which is crucial for safe circuit design.

Maximum Power Rating Constraint

Every resistor has a maximum power rating that it can safely dissipate. Exceeding this limit can cause the resistor to burn out. In circuit design, determining the maximum allowable supply voltage often involves ensuring that the power dissipated in each resistor, especially the weakest link, does not exceed its rated value.

Ohm’s Law

Ohm’s Law is a fundamental principle in circuit analysis which states that the voltage across a resistor is equal to the product of the current through it and its resistance (V = IR). This concept helps in understanding how the applied battery voltage is distributed across the resistors in a series circuit.

Series Circuit Analysis

In a series circuit, all components are connected end-to-end so that the same current flows through each one. The total resistance of the circuit is the sum of the individual resistances, and this property is used to determine the overall current and voltage distribution across the circuit components.

Voltage Division in Series Circuits

In a series circuit, the total voltage supplied by the battery is divided among the resistors in proportion to their resistances. Each resistor’s voltage drop can be calculated using the ratio of its resistance to the total resistance multiplied by the total voltage, which is essential when considering component ratings.

Transcript

00:01 So in this problem, we have three registers in series is connected by a battery of voltage v.

00:07 Register 1 is 2 -oam, register 2 -oom, and register 3 -o -on.

00:13 Also, the maximum power rating for register 1 is 4 watt.

00:18 That for the register 2 is 10 watt, and for the register 3, it is 5 -watt.

00:24 So in problem a, part a, we want to find the maximum voltage without burning up one of those registers, all right? so note that the maximum voltage in the circuit depends on the maximum current each resistance can tolerate, all right? so let's first find out the maximum current each resistors can tolerate.

00:46 So to do that, we just use pmax equals isquare max equal over r right so that means for register 1 p1 max is going to be i 1 max squared over resistance 1 so this is going to give us i 1 max is going to be square root of p1 max over r1 so this is going to be p1 max is 4 watt r1 is 2 1 which is going to be root 2 and this is going to give me 1 .4 amps.

01:39 Now similarly for i2 max it's going to be p2 max over r2 squared of that.

01:55 And this is going to give me 0 .913 amps and for i3 max it is going to be p3 max it is going to be p3 max over r3 and take the square root of that and this will give us 1 .3 amps.

02:24 Now here's the thing.

02:26 Since the register 2 is the one that has least amount of current right.

02:32 So and the register 2 is also in the series circuit so we don't want the current to be higher than 0 .913 otherwise the register 2 is going to burn and that's what we want right we don't want to burn...

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