Question

A cell is connected in series with a resistor and supplies a current of 4.0 A for a time of 500 s. During this time, 1.5 kJ of energy is dissipated in the cell and 2.5 kJ of energy is dissipated in the resistor. What is the emf of the cell

Answer 1

Answer:

$Emf=3.2\times 10^{-5}\ V$

Explanation:

Given:

Current in the circuit (I) = 4.0 A

Time for the current flow (t) = 500 s

Energy dissipated in the cell (w) = 1.5 kJ = 1500 J

Energy dissipated in the resistor (W) = 2.5 kJ = 2500 J

Emf of the cell (E) = ?

Let 'r' be the internal resistance of the cell and 'R' be the resistor of the circuit.

Now, we know that, power is equal to the product of energy dissipated and time taken.

Power dissipated in the cell is given as:

$P_{cell}=w\times t=1500\times 500=750000\ J/s$

Power dissipated in the resistor is given as:

$P_{r}=W\times t=2500\times 500=1250000\ J/s$

Now, power dissipated is also related to current and resistance as:

Power = (Current)² × Resistance.

So,

$P_{cell}=I^2r\\\\r=\frac{P_{cell}}{I^2}\\\\r=\frac{750000}{4^2}\\\\r=\frac{750000}{16}=46875\ ohm$

Similarly, resistance of the resistor is given as:

$R=\frac{P_r}{I^2}\\\\R=\frac{1250000}{16}=78125\ ohm$

Now, total resistance of the circuit is given as:

$R_{total}=R+r=78125+46875=125000\ ohm$

Now, emf of the cell is given as:

$E=\frac{I}{R_{total}}\\\\E=\frac{4}{125000}=3.2\times 10^{-5}\ V$