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Q. A $50$ volt battery is connected across a $10$ ohm resistor. The current is $4.5$ amp. The internal resistance of battery will be

Punjab PMETPunjab PMET 2001Current Electricity

Solution:

Using the relation
$i=\frac{E}{R +r}$
(given : $E=50\, V,\, R=10\, \Omega i=4.5\, amp \}$
$4.5=\frac{50}{10+r}$
$45+4.5 r=50$
$4.5\, r=50-45=5$
or $r=\frac{5}{4.5}=\frac{10}{9}=1.1\, \Omega$