# In the figure shown, the current in the 10 V battery is close to :    Option: 1 0.71 A from positive to negative terminal  Option: 2 0.42 A from positive to negative terminal Option: 3 0.21 A from positive to negative terminal   Option: 4 0.36 A from negative to positive terminal

$\begin{array}{l} \mathrm{E}_{\mathrm{eq}}=\frac{20 \times 10}{17}=\frac{200}{17} \\ \\ \text { and } \mathrm{R}_{\mathrm{eq}}=\frac{7 \times 10}{17}=\frac{70}{17} \end{array}$

\begin{aligned} &\therefore \quad I=\frac{\frac{20}{17}-10}{4+\frac{70}{17}}=0.21 \mathrm{~A}\\ &\text { from +ve to -ve terminal } \end{aligned}

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