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OpenStudy (zzr0ck3r):
that looks like Chinese to me.
OpenStudy (aaronandyson):
Okay.I'll type the question.
OpenStudy (aaronandyson):
Three resistors are connected in parallel.Draw a circuit including a switch and a battery . Derive the equivalent resistance of such a circuit.
OpenStudy (aaronandyson):
@zzr0ck3r
You there?
OpenStudy (zzr0ck3r):
nono I am saying I don't know the subject...
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OpenStudy (zzr0ck3r):
lol
OpenStudy (aaronandyson):
@welshfella @Abhisar
OpenStudy (abhisar):
Do you know anything about Ohm's law?
OpenStudy (aaronandyson):
Yep.
OpenStudy (abhisar):
Good, so lets's start. I'll leave part (a) for you.
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OpenStudy (aaronandyson):
Actually,I got part(b) also.
OpenStudy (aaronandyson):
Stcuk at part(c)
OpenStudy (aaronandyson):
Stuck*
OpenStudy (abhisar):
Ok, current in the circuit can be calculated by using the formula,
\(\sf \boxed{\Large I=\frac{Emf}{r+R}}\)
OpenStudy (abhisar):
r=internal resistance,
R=external resistance
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OpenStudy (aaronandyson):
Curent = 0.06A?
OpenStudy (aaronandyson):
0.096*
OpenStudy (abhisar):
Yes.
OpenStudy (aaronandyson):
I've no idea how to go about the rest :(
OpenStudy (abhisar):
Potential difference across each resistor:
You know the current flowing (I) now.
use V=IR for each resistor. E.g. across 20 ohms resistor voltage drop will be 0.096*20
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OpenStudy (aaronandyson):
0.96*2?
OpenStudy (abhisar):
WHat?
OpenStudy (aaronandyson):
1.92V?
OpenStudy (aaronandyson):
p.d across 20 ohm resistor is 1.92?
OpenStudy (abhisar):
YEs.
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OpenStudy (aaronandyson):
and the latter is 0.48?
OpenStudy (aaronandyson):
i.e across the 5 ohm resistor.
OpenStudy (abhisar):
Yes
OpenStudy (aaronandyson):
and across the cell?
OpenStudy (abhisar):
Ok, let's do the last one first...
When the current is flowing, then the potential difference between the terminals of the cell < EMF
and that potential drop V is given by,
V= (emf of the cell) - Current X internal resistance.
V= 2.5 - 0.096 X 1
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OpenStudy (abhisar):
Getting it?
This is the solution for last one....
OpenStudy (aaronandyson):
wait what?
OpenStudy (aaronandyson):
that should be the p.d across the cell...
OpenStudy (abhisar):
Nuuu...I explained it wrong earlier....chuck that from ur mind..
OpenStudy (abhisar):
Is it clear now???
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OpenStudy (aaronandyson):
OK!
OpenStudy (aaronandyson):
so p.d across the cell is?
OpenStudy (abhisar):
Recall what we did to find the pd across resistors???
We multiplied their individual resistances with current flowing..Right?
OpenStudy (aaronandyson):
Yep.
OpenStudy (abhisar):
Similarly, to find pd across the cell we will multiply the current with resistance of cell i.e. internal resistance...
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OpenStudy (abhisar):
Can you find it for me?/
OpenStudy (aaronandyson):
You mean 2.5*1?
OpenStudy (abhisar):
What's the current?
OpenStudy (aaronandyson):
0.096*2.5 = p.d across the cell?
OpenStudy (abhisar):
Nuu..Current X Internal resistance....
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OpenStudy (abhisar):
Current = ?
Internal resistance = ?
OpenStudy (aaronandyson):
Ugh...idk...... :'(
OpenStudy (aaronandyson):
Internal resistance = 1 ohm
OpenStudy (abhisar):
Internal resistance is given in the question and we calculated the current in 1st part :O
OpenStudy (aaronandyson):
I know...Zoned out xD
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OpenStudy (aaronandyson):
Current is 0.096A and Internal Resistance is 1 ohm
OpenStudy (aaronandyson):
Voltage drop = current*internal resitance right?
OpenStudy (aaronandyson):
resistance*
OpenStudy (abhisar):
Yes
OpenStudy (aaronandyson):
so 0.096 is the voltage drop?
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