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Question

Physics Question on Current electricity

Two batteries of emf ε1\varepsilon_{1} and ε2(ε2>ε1)\varepsilon_{2}\left(\varepsilon_{2}>\varepsilon_{1}\right)i) and internal resistances r1r_1 and r2r_2 respectively are connected in parallel as shown in figure

A

The equivalent emf εeq\varepsilon_{eq} of the two cells is between ε1\varepsilon_{1} and ε2\varepsilon_{2}, i.e. ε1<εeq<ε2\varepsilon_{1} < \varepsilon_{eq} < \varepsilon_{2}.

B

The equivalent emf εeq\varepsilon_{eq} is smaller than ε1 \varepsilon_{1}

C

The εeq\varepsilon_{eq} is given by εeq=ε1+ε2 \varepsilon_{eq} = \varepsilon_{1} +\varepsilon_{2} always

D

εeq\varepsilon_{eq} is independent of internal resistances r1r_1 and r2r_2

Answer

The equivalent emf εeq\varepsilon_{eq} of the two cells is between ε1\varepsilon_{1} and ε2\varepsilon_{2}, i.e. ε1<εeq<ε2\varepsilon_{1} < \varepsilon_{eq} < \varepsilon_{2}.

Explanation

Solution

A. The equivalent emf εeq\varepsilon_{ eq } of the two cells is between ε1\varepsilon_{1} and ε2\varepsilon_{2},