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?(A) ε = 10.0 V, r = 5.02 Ω(B) ε = 12.0 V, r = 6.00 Ω(C) ε = 4.50 V, r = 4.00 Ω(D) ε = 9.00 V, r = 2.04 Ω(E) ε = 9.00 V, r = 1.01 Ω

Answer 1

Answer:

(E) ε = 9.00 V, r = 1.01 Ω

Explanation:

As we know that power across the resistance is given as

$P = i^2 R$

now we will have

$P = 0.794 W$

$R = 100 ohm$

now we have

$0.794 = i^2(100)$

$i = \frac{V}{100 + r}$

now we can use it as

$0.794 = (\frac{V}{100 + r})^2(100)$

similarly now 100 ohm resistance is replaced by another resistance of 200 ohm

so we will have

$P = 0.401 W$

$R = 200 ohm$

now we have

$0.401 = i^2(200)$

$i = \frac{V}{200 + r}$

now we can use it as

$0.401 = (\frac{V}{200 + r})^2(200)$

now we have

$\frac{0.794}{0.401} = \frac{(200 + r)^2}{(100 + r)^2}\times \frac{100}{200}$

$1.98 = (\frac{200 + r}{100 + r})^2 \times 0.5$

$1.99 = \frac{200 + r}{100 + r}$

$199 + 1.99 r = 200 + r$

$r = 1.01 ohm$

now to find voltage of cell we will have

$V = 9 Volts$