Problem 10: Find the terminal potential difference of each cell in the circuit.
If we find the potential drop by the internal resistance and subtract it by the voltage of the emf source, we get the terminal potential of the corresponding cell. However, to find the potential drop of the internal resistance, we should know about the current in the circuit. So we apply KVL, to find the current flow in the circuit.
Assume the current flow in the clockwise direction and start from the cell of 24 V in the direction of current.
1. At the cell of 24 V, we go from the low potential (negative) to the high potential (positive). Therefore, emf is positive. 2. The internal resistance of 0.1 ohm drops potential by 0.1I with a negative sign as we go in the direction of current. 3. At the resistance of 8 ohms, we go along the direction of the current. The potential drop is 8I bearing a negative sign. 4. At the internal resistance of 0.9 ohm, the potential drop is -0.9I. 5. At the voltage source (cell) we go from positive potential to negative potential, therefore, the voltage is negative. The circuit equation is, thus,
Therefore, the current flowing in the circuit is 2 Amp. The potential drop by the internal resistance of 24 V cell is, 2*0.1 = 0.2. Therefore, use the formula to find the terminal voltage of 24 V cell = (24 – 0.2) = 23.8 V.
Similarly, the potential drop of the internal resistance of the 6 V cell is – (2*0.9) = – 1.8 V. The terminal voltage of the cell is (6+1.8) = 7.8 V. (note voltage drop is negative here).
(Try to solve the problem by applying Ohm’s law instead of Kirchhoff’s. Also consider the emfs. Hint: Find equivalent resistance of the circuit and then current I.)