184_notes:examples:week8_resistors_series

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Suppose you have the following circuit. Resistors are labeled 1 through 3 for convenience of reference. You know that the circuit contains a 12-Volt battery, and R1=10Ω, ΔV3=6 V, and the power dissipated through Resistor 1 is P1=0.1 W. What is the resistance of and power dissipated through Resistor 2?

Circuit with Resistors in Series

Facts

  • R1=10Ω
  • ΔV3=6 V
  • ΔVbat=12 V
  • P1=0.1 W

Lacking

  • R2, P2.

Approximations & Assumptions

  • The wire has very very small resistance when compared to the other resistors in the circuit.
  • The circuit is in a steady state.
  • Approximating the battery as a mechanical battery.
  • The resistors in the circuit are made of Ohmic materials.

Representations

ΔV=IR(1)

  • We represent power dissipated across a potential as

P=IΔV(2)

  • We represent the equivalent resistance of multiple resistors arranged in series as

Req=R1+R2+R3+(3)

  • We represent the Loop Rule (for potential difference within a closed loop) as

ΔV1+ΔV2+ΔV3+=0(4)

  • We represent the situation with diagram given above.

Shortly, we will constrain our calculations to just Resistors 1 and 2. We don't have any information on Resistor 2, so our approach will be to find the equivalent resistance of 1 and 2, and then focus on just Resistor 2 using equation (3). The first steps in our approach will be to find the current and potential difference across these two resistors. Note, this is not the only approach that would work! Another way would be to find individual potential difference across each resistor, and then focusing on Resistor 2 from there. (See if you can think of yet another method…)

We can use the Loop Rule – equation (4) – to find the potential difference across these two resistors. The potential difference across the battery has opposite sign as the differences across the resistors, if we consider the circuit as a loop of individual differences. We write: ΔVbat=ΔV1+ΔV2+ΔV3

Since we know ΔVbat and ΔV3, we can plug in and solve: ΔV1+ΔV2=6 V.

Plugging equation (1) into the ΔV of equation (2), we can write the power dissipated through Resistor 1 as P1=I12R1

Since we know P1 and R1, we can plug in and solve for I1=P1/R1=0.1 A. Recall that the Node Rule tells us that the current is the same everywhere in the circuit, since the entire circuit is arranged in a series. so I=I1=0.1 A.

We now have enough information to find the equivalent resistance of the two resistors, using Ohm's Law – equation (1). We write: R1 and 2, equivalent=ΔV1+ΔV2I=60Ω

Now, equation (3) tells us Req=R1+R2, so R2=ReqR1=50Ω
The power dissipated across Resistor 2 can be found using the same rewriting of equation (2) as above: P2=I2R2=0.6 W

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  • Last modified: 2017/10/16 23:22
  • by dmcpadden