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Physics 16 Online
OpenStudy (aaronandyson):

http://prntscr.com/8ie2s0

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...

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.

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

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

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.

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

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???

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...

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....

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

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?

OpenStudy (abhisar):

\(\huge \checkmark\) Voltage drop across the cell

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