physics help homework!!!?

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Question by alwaysbelieve: physics help homework!!!?
1.Three resistors of 2.00 × 103 Ω, 3.90 kΩ, and 2.00 kΩ are connected in series across a 4.00 × 102 V battery.
The voltage drop across the resistor of value 2.00 kΩ is:

2.A short piece of metal that melts when a large current passes through it is a:
a. circuit breaker c. relay
b. surge suppressor d. fuse

3.What should be the Ideal resistance of an ammeter?
a. high c. more than 20 Ω
b. medium d. zero

4.Ideally, the resistance of a voltmeter should be:
a. high c. more than 20 Ω
b. medium d. zero

Best answer:

Answer by kool-aid
2.00 × 10^3 Ω = 2kΩ

1. the equivalent resistance for a series is
2kΩ + 3.9kΩ + 2kΩ = 7.9kΩ

current through each resistor is:
V = IR
I = (4.00 × 10^2V) / 7.9kΩ = 0.0506329 A

voltage drop across the 2kΩ is:
V = IR
V = (0.0506329A)(2000Ω) = 101.27 V

2. fuse
3. zero, because it goes in series with the circuit. it measures current so you want all the current to flow through it, otherwise you’ll get inaccurate readings.
4. high, because it goes in parallel with the circuit. you want a very high resistance so that no current flows through the voltmeter and all of it stays within the circuit, resulting in better voltage reading.

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