Question

dry cell of emf 1.6 V and internal resistance 0.10 ohm is connected to a resistor of resistance R ohm. If the current drawn from the cell is 2 A, then, what will be the voltage drop across R and what is the energy dissipation in the resistor ?

Answer 1

Given,

Emf (E) = 1.6 V

Internal resistance (I) = 0.10 Ω

Current (I) = 2.0 A

As we know,

R + r = $\frac{E}{I}$

on putting the values,

R + r = $\frac{1.6}{2.0}$

R+ r = 0.8 Ω

R = 0.8 - 0.10 = 0.70 Ω

As per the question,

a) Voltage (V) drop across R ,

V = IR                       [By Ohm's law)

= 2×0.70

= 1.4 V

b) Rate of energy Dissipation inside resistor ,

energy dissipation = VI

= 1.4× 2.0

= 2.8 W.

Answer 2

According to the Question

Here we have to apply formula to get the appropriate values :-

Notation :-

E = Electro Motive force

I = Current

R = Outer resistance

r = Inner resistance

Now formula :-

⇒E = I(R + r)

Substitute the values we get :-

⇒ 1.6 = 2(R + 0.1)

⇒ R + 0.1 = 0.8

R = 0.7 ohm

Voltage drop across R :-

= Current × R

= 2 × 0.7

= 1.4 Volt

Hence Now :-

Heat dissipation :-

= I²RT

= 4 × 0.7 × T

= 2.8T J

Now lastly we have to find :-

Power dissipation = Voltage drop × current

= 1.4 × 2

= 2.8 watt