To calculate the internal resistance of the cell, we use the formula for the electromotive force (EMF) of a real cell:

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Answer

2.0 Ω\text{2.0 } \Omega

Step-by-step solution

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To calculate the internal resistance of the cell, we use the formula for the electromotive force (EMF) of a real cell:

E=I(R+r)E = I(R + r)

where EE is the EMF, II is the current, RR is the external resistance, and rr is the internal resistance.

We are given two sets of data:

  1. When R1=10.0 ΩR_1 = 10.0 \, \Omega, I1=0.5 AI_1 = 0.5 \, A
  2. When R2=22 ΩR_2 = 22 \, \Omega, I2=0.025 AI_2 = 0.025 \, A

Assumption: Using the given value I2=0.025 AI_2 = 0.025 \, A leads to a negative internal resistance, which is physically impossible. It is highly probable that I2I_2 was intended to be 0.25 A0.25 \, A. We will proceed with this assumption to find a physically meaningful answer.

  1. Write down the equations for each case using the formula E=I(R+r)E = I(R + r).

    For the first case:

    E=I1(R1+r)E = I_1(R_1 + r) E=0.5 A(10.0 Ω+r)E = 0.5 \, A (10.0 \, \Omega + r) E=5.0+0.5r(∗)E = 5.0 + 0.5r \quad (*)

    For the second case (using the assumed value I2=0.25 AI_2 = 0.25 \, A):

    E=I2(R2+r)E = I_2(R_2 + r) E=0.25 A(22 Ω+r)E = 0.25 \, A (22 \, \Omega + r) E=5.5+0.25r(∗∗)E = 5.5 + 0.25r \quad (**)

  2. Equate the two expressions for EE to solve for rr.

    5.0+0.5r=5.5+0.25r5.0 + 0.5r = 5.5 + 0.25r

  3. Rearrange the equation to solve for rr.

    0.5r−0.25r=5.5−5.00.5r - 0.25r = 5.5 - 5.0 0.25r=0.50.25r = 0.5 r=0.50.25r = \frac{0.5}{0.25} r=2 Ωr = 2 \, \Omega

  4. (Optional) Calculate the EMF EE using the value of rr to verify consistency.

    Using equation (∗)(*):

    E=5.0+0.5(2)E = 5.0 + 0.5(2) E=5.0+1.0E = 5.0 + 1.0 E=6.0 VE = 6.0 \, V

    Using equation (∗∗)(**):

    E=5.5+0.25(2)E = 5.5 + 0.25(2) E=5.5+0.5E = 5.5 + 0.5 E=6.0 VE = 6.0 \, V

    The EMF is consistent.

    The internal resistance of the cell is 2.0 Ω\text{2.0 } \Omega.

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