Physics

Physics, 25.07.2019 23:30, asims13

An 11.8 ω resistor is connected to a 10.3 v battery. what is the current in the circuit?

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Physics, 26.07.2019 23:30, emilyplays474

Resistor is connected in parallel witl a 4.0 resistor. these resistors are connected to a battery. w out, i.e, becomes an infinite resistance? ttery. itt happen if the 3. tti happen if resistor burns a) the current in the 4.0 resistor will drop to zero. b) the power dissipated in the circuit will increase g the current provided by the battery will not change d) the current in the 6.0 resistor will increase.

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Physics, 06.10.2019 01:50, rameseshajj

A100.0 resistor, a 50.0 resistor, and a 40.0 resistor are connected in parallel and placed across a 12.0 v battery. ( a. what is the equivalent resistance of the parallel circuit? incorrect: your answer is incorrect. ( b. what is the current delivered by the battery? a ( c. what is the current through each branch of the circuit? a (through 100.0 resistor) a (through 50.0 resistor) a (through 40.0 resistor)

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Physics, 02.11.2019 13:31, martl4175

1. a 15.0 kω resistor is hooked up to a 45.0 v battery in a circuit with a switch. a.) draw a circuit diagram for the circuit described. label all parts and values. b.) what is the current flowing through the resistor? c.) what is the power dissipated by the resistor? 2. a 10.0 ω resistor is hooked up in series with an 8.0 ω resistor followed by a 27.0 ω resistor. the circuit is powered by a 9.0 v battery. a.) draw a labeled circuit diagram for the circuit described. b.) calculate the equivalent resistance. c.) calculate the voltage drop across each resistor in the circuit. 3. a 12.0 v battery is hooked up with three resistors (r1, r2, r3) in parallel with resistances of 2.0 ω, 5.0 ω, and 10.0 ω, respectively. a.) draw a labeled circuit diagram for the circuit described. b.) calculate the equivalent resistance. c.) calculate the current passing through each resistor in the circuit.

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An 11.8 ω resistor is connected to a 10.3 v battery. what is the current in the circuit?...

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