Question

(e) Use the Multimeter at various points in the circuit to measure the current (i.e. between the battery and the nearest resistor, and between two resistors). NOTE: The current that you want to measure must run through the Multimeter. Therefore, attach the Multimeter in series with the circuit at the point you want to measure the current.
What is the current reading between the battery and the first resistor?

Current = ____1.90___ mA (milliamps)

What is the current reading between the first resistor and the 2nd resistor?

Current = ____1.90___ mA (milliamps)

What is the current reading between the 2nd resistor and the battery?

Current = ____1.90____ mA (milliamps)

Are all the current readings above equal to each other?

(Yes or No) Yes

2. Verify Ohm’s Law:
Ohm’s Law states that the current in a series circuit is equal to the value of battery voltage divided by the combined resistance in the circuit.

The voltage of the battery = 6.00 V (from step 1(d))

The combined resistance value = 3,040 Ohms (from step 1(c))

The quotient between these two values =

3. Power calculations for this series circuit:
(a) Record the values measured in procedures in the previous procedures:

Battery voltage = 6.00 Volts

Resistor value, R1 = 990 Ohms

Resistor value, R2 = 2,050 Ohms

Current in the series circuit = _____ mA = _______Amps


Calculate the following:
Power delivered by the battery = V I = ______ x __________
=___________ Watts


Power consumed by the resistor = R1 I2 = _________ x ( _______ )2

= __________ Watts

Power consumed by the resistor = R2 I2 = _______ x ( _________)2

= _________ Watts

Calculate how much energy would be consumed by resistor R1 in 2 hours?
(2 hours = 7200 seconds, Energy = Power x time)

Show your computation here:


Answer = _______ Joules

How many kilowatt-hours is this? (1 kWh = 3,600,000 J) (the lowest rate that the power company charges you is about $0.13 per kilowatt-hour)

Answer 1

Answer:

0.13=rs35=8500888880000