A cell supplies a current of 0.6A through 2 ohm coil and a current of 0.3A through 8 ohm coil... calculate EMF and internal resistance of the cell.(please explain also)).
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EMF is denoted with "E"
and internal resisitance with "r"
we know,
V=IR
for current 0.6 A and resistance 2ohm
V=0.6×2=1.2volt
now,
for current 0.3 A and resistance 8ohm
V=0.3×8= 2.4 volt
as we know,
V=E-Ir
so,
for V=1.2 volt
1.2=E-0.6×r......( 1 )
and
for V=2.4 volt
2.4=E-0.3 ×r........( 2 )
solving equation 1 & 2
we get
internal resistance( r ) = 4 ohm
and
EMF of cell ( E ) = 3.6 volt
hope this helps..
and internal resisitance with "r"
we know,
V=IR
for current 0.6 A and resistance 2ohm
V=0.6×2=1.2volt
now,
for current 0.3 A and resistance 8ohm
V=0.3×8= 2.4 volt
as we know,
V=E-Ir
so,
for V=1.2 volt
1.2=E-0.6×r......( 1 )
and
for V=2.4 volt
2.4=E-0.3 ×r........( 2 )
solving equation 1 & 2
we get
internal resistance( r ) = 4 ohm
and
EMF of cell ( E ) = 3.6 volt
hope this helps..
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