When an external resistance of 5 is connected across terminals of a cell, a current of 0.25 A flows through it

Question

When an external resistance of 5 $Ω$ is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 $Ω$ resistor is replaced by a 2 $Ω$ resistor, a current of 0.5 A flows through it. The internal resistance of the cell is _____ $Ω$ . [2026]

Smart Achievers · Accepted Answer

(1)

$i = \frac{ε}{R + r}$

$0.25 = \frac{ε}{5 + r}$

$0.5 = \frac{ε}{2 + r}$

$On solving, r = 1 Ω$

Solution

Q.
When an external resistance of 5 $Ω$ is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 $Ω$ resistor is replaced by a 2 $Ω$ resistor, a current of 0.5 A flows through it. The internal resistance of the cell is _____ $Ω$ . [2026]

Ans.
(1)

$i = \frac{ε}{R + r}$

$0.25 = \frac{ε}{5 + r}$

$0.5 = \frac{ε}{2 + r}$

$On solving, r = 1 Ω$

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Solution

Q. When an external resistance of 5 Ω is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 Ω resistor is replaced by a 2 Ω resistor, a current of 0.5 A flows through it. The internal resistance of the cell is _____ Ω. [2026]

Ans. (1) i=εR+r 0.25=ε5+r 0.5=ε2+r On solving, r=1 Ω

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Q.
When an external resistance of 5 $Ω$ is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 $Ω$ resistor is replaced by a 2 $Ω$ resistor, a current of 0.5 A flows through it. The internal resistance of the cell is _____ $Ω$ . [2026]

Ans.
(1)

$i = \frac{ε}{R + r}$

$0.25 = \frac{ε}{5 + r}$

$0.5 = \frac{ε}{2 + r}$

$On solving, r = 1 Ω$