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# Voltmeter reads the potential difference across the terminals of an old battery as $1.4V$ while a potentiometer reads it as $1.55V$. The voltmeter resistance is $280\; \text{ohms}$. Consider the following statements:

$\begin {array} {1 1} \text{(1) The emf of the battery is 1.4V} \\ \text{(2) The emf of the battery is 1.55V} \\ \text{(3) The internal resistance of the battery is 30 ohms} \\ \text{(4) The internal resistance of the battery is 5 ohms} \end {array}$

$\begin {array} {1 1} (A\;1\: and \: 3 & \quad (B)\;1\: and \: 4 \\ (C)\;2\: and \: 3 & \quad (D)\;2\: and \: 4 \end {array}$

The potentiometer measures the exact value of emf
(as no current flows through the battery while measuring the emf)
$1.4 = i (280)$
$\Rightarrow i = 0.005 A$
$V = E – ir$
Hence $r = 30 \: ohms$
Ans : (c)