2 cells, each of Emf 2v and internal resistance 0.5ohms are connected in series .they are made to supply current to a combination of three resistors one of the resistance 2ohms is connected in series to a parallel combination of two other resistor each of resistance 3ohms.draw the circuit diagram and Calcucate
1. Current in the circuit
2. Potential difference across the parallel combination of the resistor.
3.losts volt of the battery
Circuit Diagram:
[Cell 1] -- [2 ohms] -- [3 ohms] -- [3 ohms] -- [Cell 2]
1. Current in the circuit:
The total resistance of the circuit is given by:
R = 0.5 + 2 + (3 + 3) = 8.5 ohms
The total current in the circuit is given by:
I = E/R = 2V/8.5 = 0.235 A
2. Potential difference across the parallel combination of the resistor:
The potential difference across the parallel combination of the resistor is given by:
V = I x R = 0.235 x 6 = 1.41 V
3. Losts volt of the battery:
The losts volt of the battery is given by:
V = I x R = 0.235 x 0.5 = 0.118 V