Question

A cell has an internal resistance of 0.02ohms and e.m.f of 2.0v calculate it's terminal p.d if it's delivers (a)5A ( b)50A

Answer 1

Answer

a) 1.9 V

b) 1 V

Explanation

Let, the external resistance be R and internal resistance be r, then total voltage generated in the circuit would be

IR + Ir = E

Where, E is the e.m.f. of cell.

Now, by ohm's law we can say that IR = V, which is the potential difference between the terminals of cell.

Hence,

V + Ir = E

Here, we know r = 0.02 ohm, I = 5 A and E = 2 V

→ V + 5(0.02) = 2

→ V + 0.1 = 2

→ V = 1.9 V

Hence, the terminal p.d in first case is 1.9 V

b)

Now, the current is of 50 A

So, V + 50(0.02) = 2

→ V + 1 = 2

→ V = 1 V

Hence, the terminal p.d. now becomes 1 V

$\rule{200}2$

★ Terminal potential difference is the potential difference between the terminals of the cell.

★ It is less than EMF of the cell because the cell expends some of its energy for creating potential difference.

★ When there is no current in the circuit, the terminal potential difference = EMF of the cell

Answer 2

Answer:

$\bold\red{a)V\:=1.9}$

$\bold\green{b)V\:=1}$

______________________

Given that,

r = 0.02 ohms.

E.M. = 2 v

To calculate:-

Terminal p.d.if it delivers,a)5A b)50A

____________

IR + Ir = E

a)$\mapsto$ V + 5(0.02)=2

$\mapsto$V + 0.1 = 2

$\mapsto$ V = 2 - 01

$\mapsto$ V = 1.9

________________________

b) Now I = 50A

$\mapsto$ V +50(0.02)=2

$\mapsto$ V + 1 = 2

$\mapsto$ V = 2-1

$\mapsto$ V = 1

Hope it helps you ★ ★ ★