If 2.5V torch battery has an internal resistance of 0.7 Ω. Determine the change in terminal voltage of
the battery as the current is increased from 0 to 300mA. What value of load resistance will cause this
current to flow?
Help guys!!?
- Posted:
- 3+ months ago by Bravoboy1
- Topics:
- battery, terminal, voltage, resistance, guys, current, torch, guy
Details:
Responses (1)
We always assume perfect component values, a perfect battery, with a perfect resistance inside, and so on. In reality the resistance is in the form of bubbles on the electrode, and they move around and change over time, and sometimes depend on the amount of current.
So in this case we are assuming a perfect 2.5 volt battery in series with a perfect 0.7 ohm resistor. When current is zero the resistor drops zero volts so terminal voltage is 2.5 volts. When the current increases to 300 ma the votlage drop becomes v= IR = 300 x 10^-3A x 0.7Ω = 2.1 volts so the terminal voltage is 2.5 - 2.1 = 0.4 volts.
Load resistance at 300 ma R = E/I = 0.4V/0.3A = 1 1/3 ohms.
http://physics.nist.gov/cuu/Units/prefixes.html
You just have to memorize this stuff, at least the ones you work with all the time. It's ok to bookmark the page so you can look up the others when you run across them.
There is a prefix every three powers of ten, except for 100, 10, 1/10, and 1/100. To go from one to another you use the LARD rule: LEFT ADD RIGHT DEDUCT. When moving the decimal to the LEFT you ADD to the power of ten. When going to the RIGHT you DEDUCT from the power of ten.