Question

A cell supplies a current of 2A when connected to a

50 resistance and supplies a current of 1.2A, if
connected to a resistance of 902. Find e.m.f. and the
internal resistance r.

Answer 1

Therefore, the e.m.f. of the cell is 2556 V and the internal resistance r is 1228 Ω.

Step-by-step explanation:

Let’s use the following equations to solve this problem:

For the first case, when the cell is connected to a 50 Ω resistance and supplies a current of 2 A, we have:

I = 2 A R = 50 Ω

For the second case, when the cell is connected to a 902 Ω resistance and supplies a current of 1.2 A, we have:

I = 1.2 A R = 902 Ω

We can use the following formula to calculate the e.m.f. of the cell:

ε = I(R + r)

where ε is the e.m.f. of the cell, r is the internal resistance of the cell, and R is the external resistance.

Substituting the values from the first case, we get:

ε = 2(50 + r)

Substituting the values from the second case, we get:

ε = 1.2(902 + r)

Now we can solve for r by equating the two expressions for ε:

2(50 + r) = 1.2(902 + r)

Simplifying this equation, we get:

100 + 2r = 1082.4 + 1.2r

0.8r = 982.4

r = 1228 Ω

Now that we know the value of r, we can substitute it into either of the expressions for ε to find the e.m.f. of the cell:

ε = 2(50 + 1228) = 2556 V

Therefore, the e.m.f. of the cell is 2556 V and the internal resistance r is 1228 Ω.