The current flow in the circuit of two battery with internal resistance and two external resistance as shown in figure will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.67\,\mathrm{A} \)
Apply Kirchhoff’s voltage rule to the single loop. The positive plate (the longer line) of both cells is at the top, and the current i leaves the positive terminal of \( \varepsilon_1 \) and reaches the positive terminal of \( \varepsilon_2 \). So \( \varepsilon_2 \) opposes \( \varepsilon_1 \). All the resistances are in series, and the internal resistances are included:
\( R_{\mathrm{total}} = 1.2 + 9 + 0.8 + 4 = 15\,\Omega \)
Going round the loop in the direction of i:
\( 15 - 1.2i - 4i - 0.8i - 5 - 9i = 0 \)
\( 10 - 15i = 0 \)
\( \displaystyle i = \frac{10}{15}\,\mathrm{A} = \frac{2}{3}\,\mathrm{A} \approx 0.67\,\mathrm{A} \)