CHP35_Practice - A parallel circuit has resistances hooked...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chp. 35 Electric Circuits (SOLUTIONS) 1. T 2. T 3. F 4. T 5. F 6. A 7. C 8. B 9. C 10. A 11. A 12. C 13. B 14. 1/R equiv = 1/10Ω + 1/30Ω = 4/30Ω so : R equiv = 30/4 Ω = 7.5 Ω 15. The internal resistance of the battery is in series with the 10 Ω resistor. Ohm's Law, applied to the circuit tells us that the total resistance of the circuit can be found from: 50V = 4.5A * R total so the total resistance is: R total = 50V/4.5 = 11.11 Ω. Since the resistances are in series, they are added together to form the total and: the resistance of the battery is: 11.11 Ω - 10 Ω = .11 Ω. It's important that a car battery have a low resistance; otherwise it's liable to heat up too much and explode. 16.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A parallel circuit has resistances hooked up in multiple branches or pathways from the battery or voltage source. So, in a parallel circuit: a. The voltage is the same across each branch b. The current splits up, going through the branches in such a way, that Ohm's Law applies for each branch. The "branch currents" then join up (or, add up) on the other sides of the branches to form the original total current leaving the battery. c. From a and b, it isn't difficult to show that the total or equivalent resistance (that is, the resistance seen by the total current leaving the battery) can be found from: 1/R total = 1/R 1 + 1/R 2 + 1/R 3 + …....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online