an unknown battery (unknown emf (E) and internal resistance(r)) is connected to a 4 ohm resistor which causes a current of 2 amps to flow in the circuit. when the battery is connected to a 2 ohm resistor a current of 3 amps flows.
E-2r=8
E-3r=6
this formula may help l=E/(R+r)
where l is amps, E is emf, R is resistance in ohms and r is internal resistance
E-2r=8
E-3r=6
this formula may help l=E/(R+r)
where l is amps, E is emf, R is resistance in ohms and r is internal resistance
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3r + 6 = 2r + 8
r = 2 ohms
E = 4 + 8 = 12 volts
r = 2 ohms
E = 4 + 8 = 12 volts