When a 22 ohms resistance is connected across its terminals a potential difference 2.2V is produced. Find the internal resistance and lost volt of the cell..
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Answer:
P.D.=2.2volts , this is the EMF of the cell.
Now when the circuit is closed and a current is flowing through the circuit then the P.D.=1.8V
Thus the total resistance of the circuit is R+r=5+rohms
So, the current in the circuit is i=( 5+r 2.2
hence the PD across the internal resistance.
Given that,
5+r 2.2 )=2.2−1.8 as the drop in the potential across the battery is due to the P.D. across the internal resistance.
So, r( 5+r 2.2 )=0.4 1.8r=2
r=1.11ohms
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