Question

If you put two of these batteries in series with each other - still powering a device with the original 100 Ohm resistance - how long would the batteries last? (Note: you have two batteries now, so twice the energy to begin with than with one battery.)

Answer 1

Answer:

A certain battery can power a 100 Ohm-resistance device for 4 hours. How long can the battery power a 200 Ohm device? Ans: 8 ohms

Part B)

If you put two of these batteries in series with each other - still powering a device with the original 100 Ohm resistance - how long would the batteries last? (Note: you have two batteries now, so twice the energy to begin with than with one battery.)

t₁=8hr

t₂=2hr

Explanation:

Given that,

R₁ =100Ω

R₂ =200Ω

t₁ =4hr

Part A

The expression for energy dessipated across a resistor when it is connected with a single resistor (R) is

$E = \frac{V^2}{R} t$

Here V is the voltage across the source, t is time

energy dissipated across 200ohms is the same as  energy dissipated across 100ohms

$E = \frac{V^2}{R} t$

$\frac{V^2t_1}{R_1} = \frac{V^2t_2}{R_2}$

rearrange the above equation t₂

$t_2 = \frac{R_2t_1}{R_1} \\\\= \frac{(200)(4)}{100} \\= 8hr$

$t_2 = \frac{R_2t_1}{R_1} \\\\= \frac{(200)(4)}{100} \\= 8hr$

Part b

Now two batteries are in series

total emf =2V

$\frac{(2V)^2t_1}{R_1} = \frac{V^2t_2}{R_2} \\\\\frac{4t_1}{R_1} = \frac{t_2}{4R_2} \\\\t_2=\frac{2(4)(100)}{4(100)} \\\\=2hr$

use the equation below