Two cells A and B- each of e-m-f- 2 V- are connected in series to an external resistance R-1Omega- If the internal resistance of cell A is 1-9Omega and that of B is 0-9Omega- what is the potential difference between the terminals of cell A-

Question

Toppr Admin · Accepted Answer

Correct option is C- -0-From ohm-s law -i-dfrac-V-R-where -V- is total emf- and R is equivalent resistance-According to question-i-dfrac-2-2-1-1-9-0-9-dfrac-4-3-8- A-For cell -A- E-V-ir -Rightarrow V-2-dfrac-4-3-8-times 1-9-0-

Two cells $$A$$ and $$B$$, each of e.m.f. $$2\ V$$, are connected in series to an external resistance $$R=1\Omega$$. If the internal resistance of cell $$A$$ is $$1.9\Omega$$ and that of $$B$$ is $$0.9\Omega$$, what is the potential difference between the terminals of cell $$A$$?

Correct option is C. $$0$$
From ohm's law $$i=\dfrac{V}{R}$$
where $$V$$ is total emf, and R is equivalent resistance.
According to question:
$$i=\dfrac{2+2}{1+1.9+0.9}=\dfrac{4}{3.8}\ A$$
For cell $$A\:: E=V+ir \Rightarrow V=2-\dfrac{4}{3.8}\times 1.9=0$$.

Two cells $$A$$ and $$B$$, each of e.m.f. $$2\ V$$, are connected in series to an external resistance $$R=1\Omega$$. If the internal resistance of cell $$A$$ is $$1.9\Omega$$ and that of $$B$$ is $$0.9\Omega$$, what is the potential difference between the terminals of cell $$A$$?

Correct option is C. $$0$$From ohm's law $$i=\dfrac{V}{R}$$where $$V$$ is total emf, and R is equivalent resistance.According to question:$$i=\dfrac{2+2}{1+1.9+0.9}=\dfrac{4}{3.8}\ A$$For cell $$A\:: E=V+ir \Rightarrow V=2-\dfrac{4}{3.8}\times 1.9=0$$.

Correct option is C. $$0$$
From ohm's law $$i=\dfrac{V}{R}$$
where $$V$$ is total emf, and R is equivalent resistance.
According to question:
$$i=\dfrac{2+2}{1+1.9+0.9}=\dfrac{4}{3.8}\ A$$
For cell $$A\:: E=V+ir \Rightarrow V=2-\dfrac{4}{3.8}\times 1.9=0$$.