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: Label the circuit.
Let \( V \) be the common potential difference across the two parallel points to which R is connected. The current through R is \( I \), and the currents supplied by the two cells are \( I_1 \) and \( I_2 \).

Step 2: Apply the junction (Kirchhoff's current) rule.
The two cell currents combine to give the current in R:
\[ I = I_1 + I_2 \]

Step 3: Write terminal voltage of each cell.
For a cell of e.m.f. \( E \) and internal resistance \( r \) delivering current \( i \), the terminal voltage is \( V = E - i r \), so \( i = \dfrac{E - V}{r} \). Applying this to both cells:
\[ I_1 = \frac{E_1 - V}{r_1}, \qquad I_2 = \frac{E_2 - V}{r_2} \]

Step 4: Add the currents.
\[ I = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2} = \left(\frac{E_1}{r_1}+\frac{E_2}{r_2}\right) - V\left(\frac{1}{r_1}+\frac{1}{r_2}\right) \]

Step 5: Use the external resistor relation.
Across R, \( V = IR \). Substitute \( V = IR \):
\[ I = \left(\frac{E_1}{r_1}+\frac{E_2}{r_2}\right) - IR\left(\frac{1}{r_1}+\frac{1}{r_2}\right) \]
\[ I\left[1 + R\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\right] = \frac{E_1}{r_1}+\frac{E_2}{r_2} \]

Step 6: Solve and simplify.
Multiply numerator and denominator by \( r_1 r_2 \):
\[ I = \frac{E_1 r_2 + E_2 r_1}{r_1 r_2 + R(r_1 + r_2)} \]
This is the required current through R.
\[\boxed{\,I = \dfrac{E_1 r_2 + E_2 r_1}{r_1 r_2 + R\,(r_1 + r_2)}\,}\]