batterieschassislightningsafety
I was wondering today what could possibly happen if a lightning strikes my van where I have a 12V battery negative connected to the chassis. Could that destroy the battery or start a fire?
In order to prevent it should I disconnect the battery negative during the storm?
Thanks
Personally, I have never understood the "induced-emf" line of thinking. To me, it is easier, and much easier not to get it wrong, from the circuit and equations point of view.
I am simplifying the solenoid to be just an inductor. The equation for Circuit 1 is: $$ V_L = V_1 = L \frac{di}{dt} $$ It tells us right away that the voltage across L must be V1 and never goes to zero. The positive change of current never stops and i goes to infinity. Obviously, this is a bad model.
The obvious improvement to the model (unless the battery is underpowered) is to include the
ohmic resistance that you want to neglect. That is Circuit 2. The equation:
$$ V_1 = V_L + V_R = L \frac{di}{dt} + R i $$
Initially, i is zero. All of V1 is across L. i increases.
As i increases, voltage drops across R. Voltage across L decreases, rate of i increase slows down.
Eventually, when i reaches the level where R x i = V1. Voltage across L becomes 0.
If you solve for \$V_L\$ and \$i\$, you get something like: $$ V_L = V_1e^{-\frac{R}{L}t} $$ $$ i = \frac{V_1}{R} (1 - e^{-\frac{R}{L}t}) $$ \$V_L\$ is not the voltage across the solenoid, we need to include the voltage across R for that, the total would then be \$V_1\$.
My question is, is there not current flowing into the battery in a complete circuit?
Yes, but it is limited by the load resistor. The resistor is so-called because it resists the flow of electricity.
How come the battery is unable to handle the current under its own voltage?
The battery will have some internal resistance - in fact, we usually model them as an ideal voltage source with an internal series resistance - but the series resistance value in a good / new / healthy battery is usually very low. That means that in the short-circuit condition the current can be much higher than is safe for the battery. Think what would happen if you short out your car battery positive to the chassis with a spanner!
And if some form of resistance on the circuit is required in order to prevent a short circuit scenario, how much of a resistance do I need, in order to prevent the battery from overheating?
e.g., A 12 V car battery can supply 40 A for several minutes.
$$ R = \frac {V_{batt}}{I_{max}} = \frac {12}{40} = 0.3 \ \Omega$$
Incidentally, the power being delivered in the above example is given by \$ P = VI = 12 \cdot 40 = 480 \ \mathrm W \$.
Figure 1. (a) The apparent LED key-fob lamp. (b) The more correct model showing the cell's internal resistance. Source: LEDnique.
On a related note, the internal resistance of a button cell is used in key-fob LED lamps to limit the current to a safe value for the LED.
Best Answer
The lightning can cause a fire regardless of the battery being connected. It can go through the electronics to the battery positive as well. A lightning strike on a solid car can blow electronics, fuse metal, blow out windows and start a fire in the seats or other fabric material or said electronics.
Fire risk aside, you are relatively safer in an enclosed vehicle (No soft top or convertible) during a lightning storm. Cars have been struck while in operation with no significant damage. Others have caught on fire in the cabin, not the engine bay.
Disconnecting the battery is not important.