PHYSICS

JAMB 2001 - Question 21

Physics 2001 JAMB Past Questions - Question 21: A cell of internal resistance r supplies current to a 6.0Q resistor and its efficiency is 75%. Find the value of r

Choose the correct answers from the options given.
A cell of internal resistance r supplies current to a 6.0Q resistor and its efficiency is 75%. Find the value of r
A:
B:
C:
D:
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Correct Answer

D

Explanation

75=IR/I(R+r) *100 ;75=6/6+r *100=2ohmsThe efficiency of a device is given by the formula:

Efficiency (η) = (Useful Power Output) / (Total Power Input)

In this case, the cell with internal resistance is the device, and we want to find the internal resistance 'r'. The useful power output is the power dissipated in the external 6.0Ω resistor.

First, let's find the total power input by the cell, which is equal to the power dissipated in both the internal resistance 'r' and the 6.0Ω resistor. The power dissipated in a resistor is given by the formula:

Power = (I^2) * R

Where:
I is the current, and
R is the resistance.

The total current provided by the cell will be split between the internal resistance 'r' and the 6.0Ω resistor. Let's assume that the current supplied by the cell is 'I'. Then:

Current through internal resistance (I_internal) = I
Current through 6.0Ω resistor (I_6Ω) = I

Now, we can calculate the power dissipated in the internal resistance 'r':

Power in internal resistance = (I_internal^2) * r = (I^2) * r

And, the power dissipated in the 6.0Ω resistor:

Power in 6.0Ω resistor = (I_6Ω^2) * 6.0Ω = (I^2) * 6.0Ω

Now, we can calculate the total power input (P_in) to the cell:

P_in = Power in internal resistance + Power in 6.0Ω resistor
P_in = (I^2) * r + (I^2) * 6.0Ω

Next, we know that the efficiency (η) is given as 75%, which is 0.75 in decimal form. Therefore:

η = (Useful Power Output) / (Total Power Input)
0.75 = (Power in 6.0Ω resistor) / P_in

Now, let's find the value of P_in:

0.75 = [(I^2) * 6.0Ω] / [(I^2) * r + (I^2) * 6.0Ω]

Now, we can simplify this equation by canceling out the common factor of I^2:

0.75 = (6.0Ω) / (r + 6.0Ω)

Now, we can solve for 'r':

r + 6.0Ω = (6.0Ω) / 0.75
r + 6.0Ω = 8.0Ω

Now, subtract 6.0Ω from both sides:

r = 8.0Ω - 6.0Ω
r = 2.0Ω

So, the internal resistance 'r' is 2.0Ω.

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