Current Electricity

Question: Two resistances ${R_1}$ and ${R_2}$ are joined as shown in the figure to two batteries of e.m.f.${E_1}$ and ${E_2}$. If ${E_2}$ is short-circuited, the current through ${R_1}$is

Question Image

1) ${E_1}/{R_1}$

2) ${E_2}/{R_1}$

3) ${E_2}/{R_2}$

4) ${E_1}/({R_2} + {R_1})$

Solution:

Explanation: No Explanation

Kirchhoff's Law Cells Internal Resistance

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