Question

Eight identical cells, each of emf 2V and internal resistance 3 Ω, are connected in series to form a row. Six such rows are connected in parallel to form a battery. This battery is now connected to an external resistor R of resistance 6 Ω. Calculate:

1. emf of the battery.
2. internal resistance of the battery.
3. current flowing through R.

Answer 1

1. The emf of each cell is 2 V, and there are 8 cells connected in series. Therefore the emf of the battery is 2 V × 8 = 16 V.
2. When cells are connected in series, the total internal resistance is the sum of all the internal resistances of each cell.
   Since there are 8 cells, the total internal resistance is 3 Ω × 8 = 24 Ω.
   The formula provides the total internal resistance when connecting cells in parallel.
   `therefore 1/"R" = 1/"R"_1 + 1/"R"_2 + 1/"R"_3 + ... + 1/"R"_"L"`
   `therefore 1/"R" = 1/24 + ... + 1/24` (6 times)
   ⇒ R = 4 Ω
   Thus, the internal resistance of the battery is 4 Ω.
3. To calculate the current flowing through the external resistance R, we need to first calculate the total resistance of the circuit.
   The external resistance has a resistance of 6 Ω and the internal resistance of the battery is 4 Ω.
   When the resistors are connected in series, the total resistance is the sum of the individual resistance.
   ∴ R_total = 6 + 4 = 10 Ω
   From ohm's law
   I = `"V"/"R" = 16/10 = 1.6` A
   Therefore, the current flowing through the external resistor R is 1.6 A.