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Two batteries, one of emf 18 volts and internal resistance $2\,\Omega$ and the other of emf 12 volt and internal resistance $1\,\Omega$, are connected as shown. The voltmeter V will record a reading of
(The two batteries are connected in parallel with their like terminals joined, and the voltmeter is connected across the common terminals.)
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(The two batteries are connected in parallel with their like terminals joined, and the voltmeter is connected across the common terminals.)
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A
18 volt
B
30 volt
C
14 volt
D
15 volt
Detailed Solution
For two cells in parallel (like terminals joined), the equivalent emf is $E_{eq} = \dfrac{\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2}}{\dfrac{1}{r_1} + \dfrac{1}{r_2}} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$
An ideal voltmeter draws no current, so it reads this equivalent emf.
Here $E_1 = 18$ V, $r_1 = 2\,\Omega$, $E_2 = 12$ V, $r_2 = 1\,\Omega$.
Reading of voltmeter $= \dfrac{18 \times 1 + 12 \times 2}{2 + 1} = \dfrac{18 + 24}{3}$
$= \dfrac{42}{3} = 14$ volt
Check: the circulating current between the cells is $\dfrac{18 - 12}{2 + 1} = 2$ A, so the terminal voltage is $18 - 2 \times 2 = 14$ V $= 12 + 2 \times 1$.
An ideal voltmeter draws no current, so it reads this equivalent emf.
Here $E_1 = 18$ V, $r_1 = 2\,\Omega$, $E_2 = 12$ V, $r_2 = 1\,\Omega$.
Reading of voltmeter $= \dfrac{18 \times 1 + 12 \times 2}{2 + 1} = \dfrac{18 + 24}{3}$
$= \dfrac{42}{3} = 14$ volt
Check: the circulating current between the cells is $\dfrac{18 - 12}{2 + 1} = 2$ A, so the terminal voltage is $18 - 2 \times 2 = 14$ V $= 12 + 2 \times 1$.