Question

When a cell is connected directly across a high resistance voltmeter the reading is 1.50 V. When the cell is shorted through a low resistance ammeter, the current is 2.5 A. What are the emf and internal resistance of the cell?​

Answer 1

Answer:

this is your answer. i hope it's helpful to you

Answer 2

Answer: The emf of the cell is 1.50 V.

The internal resistance of the cell is 0.6 Ω.

To solve this problem, we can use Ohm's Law and Kirchhoff's laws to write two equations based on the two measurements:

First, when the cell is connected directly across a high resistance voltmeter, the reading is 1.50 V. This means that the voltage across the cell, V, is 1.50 V. According to Kirchhoff's voltage law, the voltage across the cell is equal to the sum of the voltage drop across the internal resistance of the cell, r, and the voltage drop across the external resistance, R (which is the resistance of the voltmeter):

                           V = IR + Ir

where I is the current in the circuit.

Second, when the cell is shorted through a low resistance ammeter, the current is 2.5 A. This means that the current in circuit, I, is 2.5 A. According to Kirchhoff's current law, the current in the circuit is the same at all points in the circuit.

We can solve these two equations for the unknowns, emf (E) and internal resistance (r):

                  V = IR + Ir

              1.50 V = 2.5 A * R + 2.5 A * r (substituting the known values)

                2.5 A = E / (R + r) (from Kirchhoff's current law)

We have two equations and two unknowns, so we can solve for E and r using algebra. First, we can solve the second equation for R + r:

                R + r = E / 2.5 A

We can then substitute this expression for R + r in the first equation:

              1.50 V = 2.5 A * (E / 2.5 A - r) + 2.5 A * r

Simplifying this equation, we get:

                 1.50 V = E - 2.5 A * r + 2.5 A * r

                1.50 V = E

Therefore, the emf of the cell is 1.50 V.

To find the internal resistance, we can substitute the value of E into one of the previous equations:

                         2.5 A = E / (R + r)

                      2.5 A = 1.50 V / (R + r)

Solving for R + r, we get:

                        R + r = 1.50 V / 2.5 A

                       R + r = 0.6 Ω

Substituting this expression for R + r into the equation we derived earlier, we can solve for r:

                     r = E / 2.5 A - R

                   r = 1.50 V / 2.5 A - 0.6 Ω

                    r = 0.6 Ω

Therefore, the internal resistance of the cell is 0.6 Ω.

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