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# circuit resistance Watch

1. 1. The problem statement, all variables and given/known data
hi
i am wondering if this is correct. this is the question;
Three cells, each of e.m.f. 1·5 V and internal resistance 0·50 ?, are connected in series to
make a battery of e.m.f. 4·5V. The battery is connected to a resistor, R, of resistance 6·0 ?.

2. Relevant equations
V= i/R

3. The attempt at a solution
TOTAL VOLTAGE = 4.5V
TOTAL RESISTANCE = 6.5 OHMS
but the answer states the total resistnace is 7.5 ohms.
can someone kindly explain how?
thanks
2. (Original post by aurao2003)
1. The problem statement, all variables and given/known data
hi
i am wondering if this is correct. this is the question;
Three cells, each of e.m.f. 1·5 V and internal resistance 0·50 ?, are connected in series to
make a battery of e.m.f. 4·5V. The battery is connected to a resistor, R, of resistance 6·0 ?.

2. Relevant equations
V= i/R

3. The attempt at a solution
TOTAL VOLTAGE = 4.5V
TOTAL RESISTANCE = 6.5 OHMS
but the answer states the total resistnace is 7.5 ohms.
can someone kindly explain how?
thanks
When components are connected in series you simply add the resistance.

Hence, 0.5+0.5+0.5+6 = 7.5 ohms
3. V = IR not I/R
4. (Original post by eswnl)
V = IR not I/R
thanks. silly me.
5. Yeah, I think you may have just misread:

"each of ... internal resistance 0·50"

So there's 3 lots of 0.5.
6. The compnents in this circuit with resistance are:

3 batteries (each of 0.5 ohms)
1 resistor (6 ohms)

(3 x 0.5) + 6 = 7.5 ohms.

Hope this helps!

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