The emfs and resistances in the given circuit have the following values.
\( E_1 = 4.2\,\mathrm{V}, E_2 = 1.9\,\mathrm{V} \)
\( r_1 = 2.0\,\Omega, r_2 = 1.6\,\Omega, R = 6.0\,\Omega \)
What is the current in the circuit?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 240 mA
The circuit is a single loop, so the same current flows everywhere, and Kirchhoff's loop rule gives it as net emf divided by total resistance.
In the figure both cells have their positive (longer) plate at the top, joined by the top wire. Going round the loop, one cell is crossed from \( - \) to \( + \) and the other from \( + \) to \( - \), so the cells oppose each other. The figure labels both internal resistances \( r_1 \); the values in the question give \( r_1 = 2.0\,\Omega \) for \( E_1 \) and \( r_2 = 1.6\,\Omega \) for \( E_2 \).
\( E_{\mathrm{net}} = E_1 - E_2 = 4.2 - 1.9 = 2.3\,\mathrm{V} \)
\( R_{\mathrm{total}} = r_1 + r_2 + R = 2.0 + 1.6 + 6.0 = 9.6\,\Omega \)
\( \displaystyle I = \frac{E_{\mathrm{net}}}{R_{\mathrm{total}}} = \frac{2.3\,\mathrm{V}}{9.6\,\Omega} = 0.24\,\mathrm{A} = 240\,\mathrm{mA} \)
The current flows clockwise, in the direction set by the larger emf \( E_1 \).