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Two cells are connected in opposition as shown. Cell $$E_1$$ is of $$8 \text{ V}$$ emf and $$2 \text{ } \Omega$$ internal resistance; the cell $$E_2$$ is of $$2 \text{ V}$$ emf and $$4 \text{ } \Omega$$ internal resistance. The terminal potential difference of cell $$E_2$$ is ______ V.
Correct Answer: 6
$$\text{Net EMF } (E_{\text{net}}) = E_1 - E_2 \implies 8 - 2 = 6\text{ V}$$
$$R_{\text{total}} = r_1 + r_2 \implies 2 + 4 = 6\ \Omega$$
$$I = \frac{E_{\text{net}}}{R_{\text{total}}} \implies \frac{6}{6} = 1\text{ A}$$
$$V_2 = E_2 + I r_2 \implies 2 + (1 \times 4) = 6\text{ V}$$
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