What kind of wire needed to prevent overheating?

[songoku]

Homework Statement

Please see below

Relevant Equations

V = I.R
$P = I^2 R$

I thought the answer is (B) because thinner wire means higher resistance and smaller current so smaller power dissipated in the wire, but the answer is (A)

What is my mistake?

Thanks

 

[haruspex]

Homework Statement:: Please see below
Relevant Equations:: V = I.R
$P = I^2 R$

View attachment 322505

I thought the answer is (B) because thinner wire means higher resistance and smaller current so smaller power dissipated in the wire, but the answer is (A)

What is my mistake?

Thanks

Where is most of the power being used? What does that tell you about how the load resistance compares to the copper wire resistance? And what does that say about the effect on the current of making the wires a bit thinner?

 

[songoku]

Where is most of the power being used?

In the heating element of the oven

What does that tell you about how the load resistance compares to the copper wire resistance?

The load resistance is higher than the copper wire resistance

And what does that say about the effect on the current of making the wires a bit thinner?

Making the wires a bit thinner will increase the resistance of the wire and also the total resistance of the circuit so the current will decrease

 

[BvU]

Making the wires a bit thinner will increase the resistance of the wire and also the total resistance of the circuit so the current will decrease

That would mean your roast isn't roasted any more (but slow-cooked instead) !
Think in terms of symbol expressions: what is the ratio of the power generated in the oven and the power generated in the connecting cable ?

$\$

 

[Steve4Physics]

I thought the answer is (B) because thinner wire means higher resistance and smaller current so smaller power dissipated in the wire, but the answer is (A)

What is my mistake?

You might like to try some simple calculations which should make it clear.

Supply voltage = 240V (since I live in the UK!)
Resistance of connecting wires = 0.1Ω
Resistance of cooker = 10Ω

Q1: Find the current.
Q2: Find the power produced in the connecting wires.
Q3 (optional): Find the power produced in the cooker.

The connecting wires are now replaced by thinner connecting wires of resistance 0.2Ω.
Q4: Find the current.
Q5: Find the power produced in the connecting wires.
Q6 (optional): Find the power produced in the cooker.

(Edited: some minor changes and I messed up the resistances!)
(Another edit: and before anyone tells me I'm wrong, yes, I know the UK is now 230V. But I'm very old.)

 

[haruspex]

Making the wires a bit thinner will increase the resistance of the wire and also the total resistance of the circuit so the current will decrease

Sure, but I asked how the two resistances compare. If you double the resistance of the copper wires, how much difference will that make to the total resistance? How much will the current drop?
Remember, the power dissipated is $I^2R$, so lowering the current a little for a large increase in resistance might increase the power lost in the wires.

 

[erobz]

Try eliminating the current before making a judgement, it's clouding the issue.

The current in the simple resistive circuit is found through:

$$V = ( r_t + R_o)I$$

And the heat dissipated in the transmission lines:

$$P_t = I^2 r_t$$

After eliminating $I$ take the derivative $\frac{dP_t}{dr_t}$; notice how it behaves for $\pm \Delta r_t$?

 

[songoku]

That would mean your roast isn't roasted any more (but slow-cooked instead) !
Think in terms of symbol expressions: what is the ratio of the power generated in the oven and the power generated in the connecting cable ?

$\$

$$\frac{P_{oven}}{P_{cable}}=\frac{R_{oven}}{R_{cable}}$$

You might like to try some simple calculations which should make it clear.

Supply voltage = 240V (since I live in the UK!)
Resistance of connecting wires = 0.1Ω
Resistance of cooker = 10Ω

Q1: Find the current.
Q2: Find the power produced in the connecting wires.
Q3 (optional): Find the power produced in the cooker.

The connecting wires are now replaced by thinner connecting wires of resistance 0.2Ω.
Q4: Find the current.
Q5: Find the power produced in the connecting wires.
Q6 (optional): Find the power produced in the cooker.

(Edited: some minor changes and I messed up the resistances!)
(Another edit: and before anyone tells me I'm wrong, yes, I know the UK is now 230V. But I'm very old.)

From calculation I got the power actually doubles when the resistance of the wire doubles.

I now understand my mistake in OP but I still can't understand the answer intuitively / qualitatively without performing calculation.

Sure, but I asked how the two resistances compare. If you double the resistance of the copper wires, how much difference will that make to the total resistance? How much will the current drop?
Remember, the power dissipated is $I^2R$, so lowering the current a little for a large increase in resistance might increase the power lost in the wires.

Doubling the resistance of the copper wires will make the total resistance to be $2R_c+R_o$ (where $R_c$ and $R_o$ are the resistance in cable and oven respectively). So the ratio between the initial total resistance and the new total resistance is $\frac{R_o + R_c}{2R_c + Ro}$

The ratio of new current to initial current will be $\frac{R_o + R_c}{2R_c + Ro}$

Try eliminating the current before making a judgement, it's clouding the issue.

The current in the simple resistive circuit is found through:

$$V = ( r_t + R_o)I$$

And the heat dissipated in the transmission lines:

$$P_t = I^2 r_t$$

After eliminating $I$ take the derivative $\frac{dP_t}{dr_t}$; notice how it behaves for $\pm \Delta r_t$?

$$P_t=\frac{V^2 ~ r_t}{(r_t+R_o)^2}$$
$$\frac{dP_t}{dr_t}=\frac{V^2 (r_t+R_o)^2-2V^2~r_t(r_t+R_o)}{(r_t+R_o)^4}$$
$$=\frac{V^2(R_o-r_t)}{(r_t+R_o)^3}$$

Then how to analyze the behavior for $\pm \Delta r_t$?

Thanks

 

[erobz]

$$P_t=\frac{V^2 ~ r_t}{(r_t+R_o)^2}$$
$$\frac{dP_t}{dr_t}=\frac{V^2 (r_t+R_o)^2-2V^2~r_t(r_t+R_o)}{(r_t+R_o)^4}$$
$$=\frac{V^2(R_o-r_t)}{(r_t+R_o)^3}$$

Then how to analyze the behavior for $\pm \Delta r_t$?

Thanks

First, we are assuming $R_o > r_t$. That implies the whole right side is always positive. Meaning if we decrease the resistance ( negative $\Delta r_t$ ) from a particular value $0<r_t<R_o$, the change in power ( $\Delta P_t$) will be in the same direction ( negative $\Delta P_t$) as the change in the resistor $r_t$ in this circuit.

 

[Steve4Physics]

From calculation I got the power actually doubles when the resistance of the wire doubles.

Well done. For ease of reference, here is the Post #5 question/answer:

We have a 240V supply and a 10Ω load.

Case 1. With thicker (0.1Ω) connecting wire:
Total resistance =10.1Ω
Current = V/R = 240V/10.1Ω = 23.8A
Power in wire = I²R = 23.8² * 0.1 = 56W

Case 2.With thinner (0.2Ω) connecting wire:
Total resistance =10.2Ω.
Current = V/R = 240V/10.2Ω = 23.5A
Power in wire = I²R = 23.5² * 0.2 = 111W

I now understand my mistake in OP but I still can't understand the answer intuitively / qualitatively without performing calculation.

See if this helps.

In Post #1 you said “thinner wire means higher resistance and smaller current”. From the above calculation we can see the current is reduced from 23.8A to 23.5A when thinner wire is used.. But that’s a pretty small change. In fact the current is approximately constant. (This happens because the load resistance is much bigger than the connecting wire resistance.)

So when you change from thicker to thinner connecting wire in this circuit, to a first approximation you can treat the current as constant. That means when the resistance of the wire increases,the power dissipated in the wire increases (P=I²R with I approximately constant.)

Note, doubling resistance halves the current only if the applied voltage is constant. But in case 1 the voltage across the connecting wires is 2.4V and in case 2 it is 4.7V.

 

[BvU]

$$\frac{P_{oven}}{P_{cable}}=\frac{R_{oven}}{R_{cable}}$$

Exactly ! Or, if you want: $$P_\text {cable} = \frac {P_\text{oven}} {R_\text{oven}} \; R_\text {cable}$$ So -- with a given power as desired in the oven, and a given $R_\text {oven}$ -- you can minimize $P_\text {cable}$ by minimizing $R_\text {cable}$

The 'intuitive way' would be: $P= I^2 R \Rightarrow P\propto R$ since $I$ is the same for the series circuit of cable and oven. You want the power in the oven and not in the cable. $R_\text {cable}=0$ would be optimum

$\$

 

[songoku]

I understand.

Thank you very much for the explanation haruspex, BvU, Steve4Physics, erobz