Question

Obtain the formula for the current flowing in the resistor R, with the help of the circuit given above. (Two cells of e.m.f. \( E_1 \) and \( E_2 \) with internal resistances \( r_1 \) and \( r_2 \) are connected in parallel and feed a common external resistor R.)

Hint:

Take a common terminal voltage \( V \) across R, write each cell current as \( (E-V)/r \), add them for the R current, then put \( V=IR \); or reduce the two parallel cells to one equivalent cell \( E_{eq},\ r_{eq} \).

Answer 1

Step 1: Replace the two cells by one equivalent cell.
Two sources in parallel behave like a single cell of effective e.m.f. \(E_{eq}\) and effective internal resistance \(r_{eq}\). This is the fastest route to the current in R.

Step 2: Equivalent internal resistance.
The internal resistances \(r_1\) and \(r_2\) are in parallel:
\[ r_{eq}=\frac{r_1 r_2}{r_1+r_2} \]

Step 3: Equivalent e.m.f.
For parallel cells the equivalent e.m.f. is the conductance-weighted average:
\[ E_{eq}=\frac{\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}}{\dfrac{1}{r_1}+\dfrac{1}{r_2}}=\frac{E_1 r_2+E_2 r_1}{r_1+r_2} \]

Step 4: Apply Ohm's law to the single-loop circuit.
The equivalent cell drives current through \(r_{eq}\) in series with R:
\[ I=\frac{E_{eq}}{R+r_{eq}} \]

Step 5: Substitute.
\[ I=\frac{\dfrac{E_1 r_2+E_2 r_1}{r_1+r_2}}{R+\dfrac{r_1 r_2}{r_1+r_2}} \]

Step 6: Clear the fractions.
Multiply top and bottom by \((r_1+r_2)\):
\[ I=\frac{E_1 r_2+E_2 r_1}{R(r_1+r_2)+r_1 r_2} \]
which matches the direct Kirchhoff result.
\[\boxed{\,I=\dfrac{E_1 r_2+E_2 r_1}{r_1 r_2+R\,(r_1+r_2)}\,}\]