When a resistor of 11Ω is connected in series with an electric cell, the current flowing in it is 0.5A . Instead when a resistor of 5Ω is connected to the same electric cell in series, the current increases by 0.4A . The internal resistance of the cell is:

Question

When a resistor of 11Ω is connected in series with an electric cell, the current flowing in it is 0.5A . Instead when a resistor of 5Ω is connected to the same electric cell in series, the current increases by 0.4A . The internal resistance of the cell is: (A) 1.5 Ω (B) 2 Ω (C) 2.5 Ω (D) 3.5 Ω

Tardigrade Admin · Accepted Answer

i=(E/R + r) 0.5=(E/11 + r) E=5.5+0.5r 0.9=(E/5 + r) E=4.5+0.9r On solving these equation, r=2.5 Ω

Q. When a resistor of $11Ω$ is connected in series with an electric cell, the current flowing in it is $0.5 A$ . Instead when a resistor of $5Ω$ is connected to the same electric cell in series, the current increases by $0.4 A$ . The internal resistance of the cell is:

$1.5 Ω$

$2 Ω$

$2.5 Ω$

$3.5 Ω$

$i = \frac{E}{R + r}$
$0.5 = \frac{E}{11 + r}$
$E = 5.5 + 0.5 r$
$0.9 = \frac{E}{5 + r}$
$E = 4.5 + 0.9 r$
On solving these equation, $r = 2.5 Ω$

Q. When a resistor of 11Ω is connected in series with an electric cell, the current flowing in it is 0.5A . Instead when a resistor of 5Ω is connected to the same electric cell in series, the current increases by 0.4A . The internal resistance of the cell is:

1.5Ω

2Ω

2.5Ω

3.5Ω