Calculating Internal Resistor in a Series Battery Circuit

[helper-]

Homework Statement

Helping someone regarding this Qns:

A Battery is formed by four 1.5 V identical cell in series. When a external resistor of 31 Ohm is connected across he battery, the current is 280 mA. Find the internal resistor of each cell.

Can someone help to solve pls?

Thank you loads (:

Homework Equations

V = E - Ir ?

The Attempt at a Solution

 

[rude man]

1. V = E - Ir ?

No, try R = E/i

 

[helper-]

R = e / i = 6V / 0.290A = 21.43 Ohms

**is this is less than the load resistance of 31 Ohms?

The answer given is 0.19 Ohms; but unable to solve. Any idea?

 

[rude man]

R = e / i = 6V / 0.290A = 21.43 Ohms

**is this is less than the load resistance of 31 Ohms?

The answer given is 0.19 Ohms; but unable to solve. Any idea?

It's 0.290 A, not 0.280.

What limits the current thru the 31 ohm resistor?

 

[truesearch]

something wrong here ! 6V emf, current of 0.28A means total R = 6/0.28 =21.4 ohms
Does not fit with external resistance of 31 ohms connected.
Check the question

 

[rude man]

something wrong here ! 6V emf, current of 0.28A means total R = 6/0.28 =21.4 ohms
Does not fit with external resistance of 31 ohms connected.
Check the question

That's right, it's impossible.

 

[helper-]

*opps type on the 0.28.

Thanks all! Guess have to check with the lecturer, seems the qns is wierdddd. (: