You would like to check if the battery voltage drops while it is supplying a current of 6A. You use a voltmeter designed to measure voltages up to 20V and having a resistance of 50,000 Ohms.
This connection is in series so V= I (R1+R2)
V / (R1+R2) = I
12 / (2+ 50000) = I
2.39E-4 = I
How much current (in A) would flow through the headlight for Circuit b?
The connection is in parallel so Req = (1/ R1 + 1/R2)
Req = (1/2 + 1/50000) = .50002
I = V / R
I = 12 / .50002
I want to check if what I have done is right.
The Attempt at a Solution
Is right bellow the questions.