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Current Divider Circuits
10 questions By Tony R. Kuphaldt
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Question 7 of 10
Calculate the percentage of total current for each resistor in this parallel circuit:
Reveal answerR1 = 50.3% of total current
R2 = 27.6% of total current
R3 = 22.1% of total current
Notes:Nothing to comment on here, really. Just a straight-forward current divider formula problem!
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Question 8 of 10
Calculate the proper value of resistance R2 needs to be in order to draw 40% of the total current in this circuit:
Reveal answerR2 = 1.5 kΩ
Follow-up question: explain how you could arrive at a rough estimate of R2‘s necessary value without doing any algebra. In other words, show how you could at least set limits on R2‘s value (i.e. “We know it has to be less than . . .” or “We know it has to be greater than . . .”).
Notes:This is an interesting problem to solve algebraically from the current divider formula. I recommend using the product-over-sum formula for parallel resistance if you plan on doing this algebraically. The estimation question (in the follow-up) is also very good to discuss with your students. It is possible to at least “bracket” the value of R2 between two different resistance values without doing any math more complex than simple (fractional) arithmetic.
Of course, a less refined approach to solving this problem would be to assume a certain battery voltage and work with numerical figures - but what fun is that?
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Question 9 of 10
A student is trying to use the “current divider formula” to calculate current through the second light bulb in a three-lamp lighting circuit (typical for an American household):
The student uses Joule’s Law to calculate the resistance of each lamp (240 Ω), and uses the parallel resistance formula to calculate the circuit’s total resistance (80 Ω). With the latter figure, the student also calculates the circuit’s total (source) current: 1.5 A.
Plugging this into the current divider formula, the current through any one lamp turns out to be:
I = Itotal ( Rtotal R) = 1.5 A ( 80 Ω 240 Ω) = 0.5 A This value of 0.5 amps per light bulb correlates with the value obtained from Joule’s Law directly for each lamp: 0.5 amps from the given values of 120 volts and 60 watts.
The trouble is, something doesn’t add up when the student re-calculates for a scenario where one of the switches is open:
With only two light bulbs in operation, the student knows the total resistance must be different than before: 120 Ω instead of 80 Ω. However, when the student plugs these figures into the current divider formula, the result seems to conflict with what Joule’s Law predicts for each lamp’s current draw:
I = Itotal ( Rtotal R) = 1.5 A ( 120 Ω 240 Ω) = 0.75 A At 0.75 amps per light bulb, the wattage is no longer 60 W. According to Joule’s Law, it will now be 90 watts (120 volts at 0.75 amps). What is wrong here? Where did the student make a mistake?
Reveal answerThe student incorrectly assumed that total current in the circuit would remain unchanged after the switch opened. By the way, this is a very common conceptual misunderstanding among new students as they learn about parallel circuits!
Notes:I am surprised how often this principle is misunderstood by students as they first learn about parallel circuits. It seems natural for many of them to assume that total circuit current is a constant when the source is actually a constant-voltage source!
