A 10   V battery with internal resistance 1   Ω and a 15   V battery with internal…

A $10\text{V}$ battery with internal resistance $\text{1}\mathrm{\Omega }$ and a $15\text{V}$ battery with internal resistance $\mathrm{0.6 }\mathrm{\Omega }$ are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to:

1. $11.9 \mathrm{V}$
2. $13.1 \mathrm{V}$
3. $12.5 \mathrm{V}$
4. $24.5 \mathrm{V}$

Solution

The equivalent emf of the battery parallel combination is given as

Equation $= \frac{\frac{E_1}{r_1 }+ \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$

$= \frac{\frac{10}{1} + \frac{15}{0.6}}{\frac{1 }{1} + \frac{1}{ 0.6}}= \frac{10+ \frac{150}{6}}{1+ \frac{10}{6}}$

$=\frac{105}{8}$

$=13.1 \mathrm{V}$

$\therefore$ The reading measured by voltmeter $=13.1 \mathrm{V}$

Asked in: JEE Main 2015 (10 Apr Online)

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