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=v/i, 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 51.88, 91.88, and 26.25. by doing this, my amp went to 0.05A and i still have 9 V.

Resistance for components of a circuit in series is additive. That is, for two resistors R1 and R2 in series, the total resistance RtotR1 R2, or for any n resistors Rtot4:1 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. Circuit Grab Bag 3.00 V Visual OLifelike Schematic Show Values Wire Move over a wire to read current Tools Voltmeter Resistor Ammeter(s) Non-Contact , Ammeter Size O Large o Medium O Small Advanced Battery Light Bulb Show Reset AlI Switch Help!

Explanation / Answer

given three resistors R1,R2,R3 are connected in series across a battery

so total resistance of the circuit is R =R1+R2+R3

let the emf of voltage is V

total current in the circuit is=V/(R1+R2+R3)

since the resistors are connected in series same current pass through each resistor is V/(R1+R2+R3)

voltage across resistor R1is =VR1/(R1+R2+R3)

voltage across resistor R2 is=VR2/(R1+R2+R3)

voltage across resistor R3 is=VR3/(R1+R2+R3)

since resistors are connected in series the total resistance of the circuit is equal to sum of the individual resistances

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.