If a cell of emf 2V and internal resistance 0.5 ohm is connected across a resistance R, the current that flows is the same as that when a cell of emf 1.5 V and internal resistance 0.3 ohms is connected across the same resistor. Then R= Ohm.

Explanation

Figure 1 describes the first cell and figure 2 describes the second cell, both carrying the same current i. i is the circuit current and to find R Using ohm's law for figure 1. V=IR V= i Req1 , where i is the circuit current and Req1 is the resistance of the first cell. Req1 = 0.5 + R (since they are in series the resistance are added) V= i Req1 2= i (0.5 + R) i = 2 / 0.5 + R - eq1 Using ohm's law for figure 2. V=IR V= i Req2I, where i is the circuit current, and Req2 is the resistance of the second cell. Req = 0.3 + R (since they are in series the resistance are added) V= i Req2 1.5 = i (0.3 + R) i= 1.5 / 0.3 + R -- eq2 Equating both eq1 and eq2 2 / 0.5 + R = .5 / 0.3 + R 0.6 + 2R = 0.75 + 1.5R 0.5R = 0.15 R = 0.15/0.5 R="0.3 ohms"