Physics 2007 JAMB Past Questions - Question 17: three cells each of emf 1.5v and internal resistance 2.5v are connected in parallel .find the net emf and the internal resistance

Correct Answer

B

Explanation

When three cells with an electromotive force (EMF) of 1.5 volts and an internal resistance of 2.5 ohms are connected in parallel, you can calculate the net EMF and internal resistance as follows:

1. Net EMF (E_net) in parallel connection remains the same as the EMF of a single cell. So, in this case, E_net is 1.5 volts.

2. To calculate the internal resistance (R_internal) in parallel, you can use the formula:

  \[ \frac{1}{R_internal} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

  Where:
  - \(R_1\) is the internal resistance of the first cell (2.5 ohms).
  - \(R_2\) is the internal resistance of the second cell (2.5 ohms).
  - \(R_3\) is the internal resistance of the third cell (2.5 ohms).

  Substituting these values into the formula:

  \[ \frac{1}{R_internal} = \frac{1}{2.5} + \frac{1}{2.5} + \frac{1}{2.5} \]

  \[ \frac{1}{R_internal} = \frac{3}{2.5} \]

  Now, calculate the reciprocal to find \(R_internal\):

  \[ R_internal = \frac{2.5}{3} \approx 0.833 \text{ ohms} \]

So, the net EMF of the parallel connection is 1.5 volts, and the internal resistance is approximately 0.833 ohms.