An electric generator has e.m.f. of 240 V and an internal resistance of 1Ω if the current supplied by the generator is 20A when the terminal voltage is 220 V, find the ratio of the power supplied to the power dissipated?
11:1
1:11
12:11
11:12
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Option A is correct
P = IV = V²/R
EMF = 220V
For Supplied = 240V
Then, 240-220=20V for dissipated
P=V²/R = 20²/1 =400watts
Then, for supplied
P= IV = 220X20=4400watts
Supplied: dissipated
4400:400
44:4
11:1
@myschool answer is c because;
Power supplies =IE
Power dissipated=IV
Power supplied:power dissipated
IE/IV=E/V
240/220=12/11✔️✔️💯
Ref:All inclusive calculation in physics
Power supplied = I²R = 20²×R
Given the e.m.f to be 240v
And E = IR
Then, 240 = 20×R
R = 12.
Power supplied = 20²×12 =4800wat
Power dissipated = I²R
Where E=IR and we're given the terminal voltage to be 220v
Then E=IR
220 = 20 × R
R= 11.
Power dissipated = I²R = 20²×11 = 4400wat
Ratio = 4800watt/4400watt = 12/11
And again, just to add to what have been said;
The e.m.f. (240 V) was not used directly to calculate the power supplied to the external circuit because the e.m.f. is the total voltage the generator can produce. However, some of that voltage gets "used up" overcoming the internal resistance within the generator itself.
Let us think about it in this way:
Imagine you have a water pump that can push water with a total pressure (that's like the e.m.f.). But there's a little bit of friction inside the pump (that's like the internal resistance). So, the water that actually comes out of the pump's outlet has a slightly lower pressure (that's like the terminal voltage) because some of the pressure was used to overcome the internal friction.
The power supplied to the things connected to the generator depends on the actual voltage they receive, which is the terminal voltage (220 V in this case), not the total potential the generator can create (the e.m.f.).
The 20A current supplied by the generator passes through its 1R internal resistance, and that resistance dissipates I^2 * R = 20^2 * 1 = 400 Watts.
The voltage drop across that resistance, at 20A load = I * R = 20 * 1 = 20 Volts. hence the generator’s unloaded voltage of 240V drops to 220V at 20A load.
If you’re asking the ratio of the power supplied TO THE EXTERNAL LOAD to the power dissipated IN THE GENERATOR’s INTERNAL RESISTANCE, that ratio is 220V * 20A = 4,400W to 400W or 11:1.
1/12 of the power generated is dissipated in its internal resistance and 11/12 is dissipated in the load resistance.
Culled from Quora.
we are trying to find the ratio of power suppled to power dissipated so use the supplied voltage 20×1=20V and dissipated voltage=220V tgat is 20/220.
no need to solve for others provided they are all constant the final answer is 20/220= 1/11
AGADIDO, i gatch u
The formula for power supplied is:
P=IV.
While the formula for power dissipated is:
P=I^2R.
I think that's where the confusion is coming from.
To find the ratio of the power supplied to the power dissipated, we can use the formula P = IV, where P is power, I is current, and V is voltage.
The power supplied by the generator is given by P_supplied = IV, where I is the current and V is the terminal voltage.
The power dissipated in the internal resistance is given by P_dissipated = I^2 * R, where I is the current and R is the internal resistance.
In this case, the current is 20A, the terminal voltage is 220V, and the internal resistance is 1Ω.
So, the power supplied is P_supplied = 20A * 220V = 4400W.
And the power dissipated is P_dissipated = (20A)^2 * 1Ω = 400W.
To find the ratio, we divide the power supplied by the power dissipated:
Ratio = P_supplied / P_dissipated = 4400W / 400W = 11:1.
So the ratio of the power supplied to the power dissipated is 11:1.
The answer is A. 11:1.
