an electric cell of internal resistance of 0.5Π deliver a a current of 2.0A when resistance of 3Πis connected across it. the emf of the cell is
emf=I(Rint+R)=2.0*(.5+3)=7volts
To find the emf (electromotive force) of the cell, we can use Ohm's law and the concept of internal resistance.
1. First, we need to understand the relationship between the total current, internal resistance, and external resistance in a circuit. According to Ohm's law:
V = I * R
Where V is the voltage across the circuit (emf + voltage drop due to the internal resistance), I is the total current flowing through the circuit, and R is the total resistance.
2. In this case, the total current is given as 2.0A and the external resistance is 3Π.
3. The internal resistance of the cell is given as 0.5Π.
4. In order to find the emf, we need to isolate the emf term in Ohm's law equation. Rearranging the equation, we have:
V = I * R + I * r
Where:
- V is the voltage across the circuit (emf + voltage drop due to the internal resistance)
- I is the total current flowing through the circuit
- R is the external resistance
- r is the internal resistance
5. Substituting the given values:
V = 2.0A * 3Π + 2.0A * 0.5Π
6. Simplifying the equation:
V = 6Π + Π
V = 7Π (emf of the cell)
Therefore, the emf of the cell is 7Π.