There may be a trick to simplify it because of the symmetry of the resistance values, but it's probably just easiest to write the KCL equations and solve them. Can you show us that work?Vanessa Avila said:Homework Statement
I am confused what to do but I have to find the current through the battery:
Homework Equations
V=IR
The Attempt at a Solution
I decided to find the equivalent resistance and do V/R = I using the 14V, but I believe my Req is wrong. What is throwing me off is the horizontal 1 ohm resistor.
You'll have to show us what ##i_1## is on your diagram, but I would say that it's very unlikely. Any current through R3 would "break" any attempt to consider the vertical paths as isolated series connected resistors.Vanessa Avila said:
My bad. Here's what I was trying to do:gneill said:
You can either write the multiple KVL equations, or the multiple KCL equations. I personally prefer KCL, but either will work. Are you familiar with how to write them?Vanessa Avila said:My bad. Here's what I was trying to do:
I realized that the I1 splits as soon as it reaches the junction and i can either keep going to the right or down. So that means my equation for my loop has to involve I2 right?
Yes. How my professor taught it to us is that we have to set directions for the current and create a clockwise or a counter-clockwise loop. My biggest problem with it is keeping it to only 3 unknown currents (I1, I2, and I3) as he advised. I get confused when I am past assigning where I3 is.berkeman said:
gneill said:
would it be -I3-I2?gneill said:
No. remember, whatever is flowing out of the node must also be flowing into the node. You have I2 and I3 both flowing out, so what must flow in on the remaining path?Vanessa Avila said:
Yup.Vanessa Avila said:
No worries. After a bit of practice it'll suddenly "cick" and you'll wonder what all the fuss was aboutSorry, I'm a little confused about this stuff
Oh okay! Thanksgneill said:Yup.
No worries. After a bit of practice it'll suddenly "cick" and you'll wonder what all the fuss was about
:) Awesome! So finding the current through the battery, i have to find I1, I2 and I3 and sum them together to use in the V=IR equation, where V = 14V?gneill said:
Your I1 flows through the battery. Find I1 and you're done (of course you need to mostly solve everything to get there... unless you use something like Cramer's Rule to find just the one current).Vanessa Avila said:
gneill said:
To calculate the current through a battery, you can use Ohm's Law which states that current (I) is equal to voltage (V) divided by resistance (R). So, I = V/R. Make sure to use the correct units for voltage (volts) and resistance (ohms).
The current through a battery is affected by the voltage of the battery, the resistance of the circuit, and the internal resistance of the battery. Other factors such as temperature and the type of battery can also have an impact on the current.
No, the current through a battery is not always constant. It can vary depending on the resistance of the circuit and the voltage of the battery. In some cases, the internal resistance of the battery can also cause fluctuations in the current.
Yes, the current through a battery can be too high. This can happen if the resistance of the circuit is too low or if the battery is producing a higher voltage than what the circuit can handle. High current can also lead to overheating and potential damage to the circuit or battery.
To measure the current through a battery, you can use a multimeter. Make sure to set the multimeter to measure current in the appropriate range and connect it in series with the battery and the circuit. Another option is to use a current sensor that can be connected to a data logger or a computer for more precise measurements.