Each month, College of Science alumni have the opportunity to test their skills and solve a current science student exam question. These questions are posed in the monthly College of Science newsletter, Discovery Monthly. The very first Science Question of the Month was featured in the 2022 alumni magazine Discovery.
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The January Science Question of the Month, provided by physics professor Kristin Lewis, was pulled from a Physics 181 exam.
Question
Calculate the current.
Consider the direct current (DC) circuit pictured, which contains one battery of terminal voltage DV = 11 Volts [V], and three different resistors, each with resistance measured in Ohms [W] and proportional to resistance R.
The circuit starts at the positive terminal of a battery labeled "delta V", extends up to a resistor labeled 3R, and extends to the right to split into two parallel vertical branches. The branch on the left has a resistor labeled R. The branch on the right has a resistor labeled 2R . The branches recombine, and the circuit ends at the negative terminal of the battery.
What is the current through the resistor of resistance R? Your answer should be in units of Amperes [A], and it will be a function of R.
Solution
Note the following relationship between units: [1 Ampere] = [1 Volt]/[1 Ohm]. Units have been suppressed in the solution.
Step 1: Find the equivalent resistance of the combination of resistors. First, combine the two resistors in parallel: 2R and R. Label the equivalent resistance of this combination as RR2R, and sketch the “intermediate” circuit.
(1/R)+(1/2R)=(1/RR2R)
Therefore,
RR2R=(2R/3)
Next, combine the two remaining resistors, which are in series. Label the equivalent resistance, which is the equivalent resistance of the entire circuit, Req.
Req=3R+RR2R=3R+(2R/3)=(11R/3)
Step 2: Find the current drawn from the battery. The potential difference across Req is equal to the potential difference across the terminals of the battery, and the potential difference across any resistor is equal to the current through it times its resistance (â–²V=IR, *important relationship*). Therefore, the current drawn from the battery of terminal voltage 11 V is
&²Ô²ú²õ±è;±õ=(â–²V/¸éeq)=(11/[11R/3])=(3/R)
Step 3: Consider this current, I, through the “intermediate circuit”. Because the two resistors are in series the current through each is the same and equal to I=(3/R). Solve for the potential difference across RR2R:
(Check: â–²V3R=I(3R)=(3/R)(3R)=9, â–²V3R+â–²VR2R=9+2=11=â–²V)
&²Ô²ú²õ±è;â–²VR2R=I(RR2R)=(3/R)(2R/3)=2
Step 4: Consider the original circuit. The potential difference across resistors in parallel is the same. Therefore, the potential difference across resistors R and 2R is equal to that across their equivalent resistor RR2R: â–²VR=&²Ô²ú²õ±è;â–²V2R =&²Ô²ú²õ±è;â–²VR2R =2. Knowing the potential difference across R, solve for the current through it, IR:
IR = (/R)=(2/R)
(Check: I2R=°Úâ–²V2R/2R]=[2/2R]=[1/R], IR+I2R=[2/R]+[1/R]=[3/R]=I)
Answer:
IR = 2/R (units: A)
January's Winner
January's winner was Gunes Kaplan, who was first to submit the correct response. Congratulations, Gunes Kaplan!
Kaplan graduated from the ÇéÉ«ÎåÔÂÌì, Reno with her master's degree (2005) and doctoral degree (2017) in physics. Kaplan grew up in Turkiye (formerly known as Turkey). She has been working for the State Department of Education in Carson City since 2013. She is shown here with her miniature Australian Shepherd, Prince. Kaplan recently wrote a NSights blog post about the earthquakes in Turkey and Syria.
