help me w this phys q(:?

Question

a battery is connected in series with a 2 ohms resistor and a switch. a voltmeter connected across the battery reads 12 V when the switch is open but 8 when it is closed. what is the internal resistance of the battery?

Answer 1

Actual voltage of the battery V=12v , now when the key is closed there is loss of voltage across the battery do to flow of current against its internal resistence. If R (int) is the internal resistence and R (se) is the series resistence. Therefore
V- IXR(int)= 8 volt, or 12- IXR(int)= 8 volt, Therefore IXR(int) = 4volt. Now current through the circuit when the key is closed is I= V/(R(int) + R(se))= 12/(R(int)+2)
Therefore IXR(int)=(12/(R(int)+2))XR(int)
or, 4= (12/(R(int)+2))XR(int)
Solving we get R(int)=1 ohm.

Answer 2

The voltage source is V and we can write
V= I(R1+R2) where R2 is battery's internal resistance
The current I is the same in both resisstors
I= V/(R1+R2) then
Internal voltage drop across internal resistor is
V2= [V/(R1+R2)]R2
then
V1 (R1 +R2)=V R1
R2= (V R1 - V1R1)/V1
R2= (12 x 2 - 8 x 2)/ 8
R2=1 ohm

Answer 3

8 = 2*J , get J
12-8 = J*r , get r (internal resistance)

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