When a battery connected across a resistor of 16 Ω, voltage across the resistor is 12 V. When the same battery is connected across a resistor of 10 Ω, voltage across it is 11 V. The internal resistance of the battery (in ohm) is

Question

When a battery connected across a resistor of 16 Ω, voltage across the resistor is 12 V. When the same battery is connected across a resistor of 10 Ω, voltage across it is 11 V. The internal resistance of the battery (in ohm) is (A)  (10/7)  (B)  (20/7)  (C)  (25/7)  (D)  (30/7)

Tardigrade Admin · Accepted Answer

Here, V < E ∴ E=V+I r For first case, E=12+(12/16) r ...(i) For second case, E=11+(11/10) r ...(ii) From Eqs. (i) and (ii), we get 12+(12/16) r =11+(11/100) r ⇒ r =(20/7) Ω

Q. When a battery connected across a resistor of $16 Ω$ , voltage across the resistor is $12 V$ . When the same battery is connected across a resistor of $10 Ω$ , voltage across it is $11 V$ . The internal resistance of the battery (in ohm) is

$\frac{10}{7}$

$\frac{20}{7}$

$\frac{25}{7}$

$\frac{30}{7}$

Here, $V < E$
$∴ E = V + I r$
For first case,
$E = 12 + \frac{12}{16} r ... (i)$
For second case,
$E = 11 + \frac{11}{10} r ... (ii)$
From Eqs. (i) and (ii), we get
$12 + \frac{12}{16} r = 11 + \frac{11}{100} r$
$\Rightarrow r = \frac{20}{7} Ω$

Q. When a battery connected across a resistor of 16Ω, voltage across the resistor is 12V. When the same battery is connected across a resistor of 10Ω, voltage across it is 11V. The internal resistance of the battery (in ohm) is

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