Question

For the circuit given below choose the correct options, (Given: 'r' is the internal resistance of the cell connected to an external resistor 'R')

• A.
  
• B.
  
• C.
  
• D. All of these

Answer 1

The correct option is D

All of these

When 'n' cells of emf 'E' are connected in series then the equivalent emf of the circuit is 'n E'. SO, here the cells have internal resistance 'r' and emf 'E'. Therefore, the equivalent resistance and the emf becomes 'R + 3r' and '3E' respectively. Now, in series the current through every resistance is same, so current through every resistor in the circuit becomes $\frac{3E}{3r+R}$. As we know that the potential difference across any resistor is the product of its resistance and currrent flowing(ohm's law). SO, $V=\frac{3E}{3r+R}.xR$ and since we know Power developed in any resistor is the product of the potential difference across the resisitor and the current flowing through it. Therefore,$P={\left(\frac{3E}{3r+R}\right)}^{2}R$