Hi, i need some help filling the chart data. The first part where the leads are

Question

Hi, i need some help filling the chart data. The first part where the leads are on a resitor, i have 3 volts and 0.30 amp, using ohms law for R=wi, i have 10. then when i moved the leads to the battery, i have 9 volts and 0.30 amps, which= 30. On the last step i have changed the ohms on the 3 resistors to 5188, 9188, and 26.25. by doing this, my amp went to 0.05A and i still have 9 V instruction for lab is below Resistance for components of a circuit in series is additive. That is, for two resistors R1 and R2 in series, the total resistance R -R1 + R2, or for any n resistors R 1-1 Ri. In the construction kit set up a circuit with three resistors and a battery in series. In the toolbox on the right check the boxes for the voltmeter and the non-contact ammeter tot tot Circuit Grab Bag 00 V Visual o Litelike Schematic | | Show values Wire to read current Tools Non-contact Ammeter Battery Size a Medium Small Light BubAdvanced Show Reset AllI

Explanation / Answer

As we know from the reading of the voltmeter that the potential drop across the resistor is
3 V.
We know in the series current remain same and potential will be added up.
Hence total potential that will be applied by the battery equal to the potential drop in all resistor that is
= 3 + 3+ 3 = 9 V
therefore the voltage applied to the circuit by battery is 9V.
Now from the ammeter reading we know that the current is 0.3 A.
hence in series current will be same in each resistor.
Appying ohm law in first resistor
V = IR
3 = 0.3*R
R = 10 ohm
Similalry other two resistor will be 10 and 10 ohm.
Now we can fill up the table.
Potential drop in each resistor = 3V
Potential drop in circuit = 9V
Current in each resistor = 0.3 A
Current in circuit = 0.3 A
Resistance of each resistor = 10 ohm
Resistance of circuit = 30 ohm

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