Answer:
21.6 ohm
Explanation:
We are given that
EMF=E=9.63 V
Current=I=118 mA=[tex]118\times 10^{-3} A[/tex]
[tex]1 mA=10^{-3} A[/tex]
Resistance=[tex]R=60\Omega[/tex]
We have to find the internal resistance of the battery.
We know that
[tex]V=E-Ir[/tex]
We know that V=IR
[tex]IR=E-Ir[/tex]
[tex]IR+Ir=E[/tex]
[tex]I(R+r)=E[/tex]
[tex]R+r=\frac{E}{I}[/tex]
Substitute the values
[tex]60+r=\frac{9.63}{118\times 10^{-3}}[/tex]
[tex]60+r=81.6[/tex]
[tex]r=81.6-60[/tex]
[tex]r=21.6\Omega[/tex]
Hence, the internal resistance of the battery=21.6 ohm
Answer:
21.6 ohm
Explanation:
EMF of the battery, E = 9.63 V
Current, i = 118 mA = 0.118 A
Resistance, R = 60 ohm
Let the internal resistance of the cell is r.
[tex]i = \frac{E}{R + r}[/tex]
R + r = 9.63 / 0.118
60 + r = 81.6
r = 21.6 ohm
