Question

A parallel combination, as stated in (a) above, of two cells of emfs \(E\) and \(3E\) and internal resistances \(R\) each is connected across a resistance \(2R\). Find the current that flows through resistance \(2R\).

Hint:

For cells connected in parallel: \[ E_{\text{eq}} = \frac{E_1r_2+E_2r_1}{r_1+r_2} \] \[ r_{\text{eq}} = \frac{r_1r_2}{r_1+r_2} \] Always reduce the combination to a single equivalent cell before applying Ohm's law.

Answer 1

Concept: The two cells are first replaced by their equivalent cell. Once the equivalent emf and equivalent internal resistance are obtained, the circuit becomes a simple series combination of:
• equivalent emf \(E_{\text{eq}}\),
• equivalent internal resistance \(r_{\text{eq}}\),
• external resistance \(2R\). The current can then be found using Ohm's law.

Step 1: Write the given data. For the first cell, \[ E_1=E, \qquad r_1=R. \] For the second cell, \[ E_2=3E, \qquad r_2=R. \]

Step 2: Calculate the equivalent emf. Using \[ E_{\text{eq}} = \frac{E_1r_2+E_2r_1} {r_1+r_2}, \] we get \[ E_{\text{eq}} = \frac{(E)(R)+(3E)(R)} {R+R}. \] \[ E_{\text{eq}} = \frac{4ER}{2R}. \] \[ \boxed{ E_{\text{eq}}=2E } \]

Step 3: Calculate the equivalent internal resistance. Using \[ r_{\text{eq}} = \frac{r_1r_2} {r_1+r_2}, \] \[ r_{\text{eq}} = \frac{R\times R} {R+R}. \] \[ r_{\text{eq}} = \frac{R^2}{2R}. \] \[ \boxed{ r_{\text{eq}} = \frac{R}{2} } \]

Step 4: Determine total circuit resistance. External resistance \[ =2R. \] Total resistance in the circuit is \[ R_{\text{total}} = 2R+\frac{R}{2}. \] Taking LCM, \[ R_{\text{total}} = \frac{4R+R}{2}. \] \[ R_{\text{total}} = \frac{5R}{2}. \]

Step 5: Calculate the current. Using Ohm's law, \[ I = \frac{E_{\text{eq}}} {R_{\text{total}}}. \] Substituting, \[ I = \frac{2E} {\frac{5R}{2}}. \] \[ I = 2E\times\frac{2}{5R}. \] \[ \boxed{ I = \frac{4E}{5R} } \] Final Answer: The current through the resistance \(2R\) is \[ \boxed{ I=\frac{4E}{5R} } \]