Question

When a 100-Ω resistor is connected across the terminals of a battery of emf ε and internal resistance r, the battery delivers 0.794 W of power to the 100-Ω resistor. When the 100-Ω resistor is replaced by a 200-Ω resistor, the battery delivers 0.401 W of power to the 200-Ω resistor. What are the emf and internal resistance of the battery?

Answer 1

Answer:

The emf and internal resistance of the battery are 9 volt and 2.04 Ω

Explanation:

Given that,

First resistor$R = 100\ \Omega$

Power$P = 0.794\ W$

Second resistor$R=200\ \Omega$

Power$P=0.401\ W$

We need to calculate the emf and internal resistance

Using formula of emf

$P=\dfrac{e^2}{R+r}$

For first resistor

$0.794=\dfrac{e^2}{100+r}$....(I)

For second resistor

$0.401=\dfrac{e^2}{200+r}$.....(II)

From equation (I) and (II)

$\dfrac{795}{401}=\dfrac{200+r}{100+r}$

Using component of dividend rule

$\dfrac{393}{401}=\dfrac{100}{100+r}$

$r =2.04\ \Omega$

Now, Put the value of internal resistance in equation (I)

$0.794=\dfrac{e^2}{100+2.04}$

$e^2=0.794(100+2.04)$

$e=\sqrt{0.794(100+2.04)}$

$e=9\ volt$

Hence, The emf and internal resistance of the battery are 9 volt and 2.04 Ω