ChatterBank13 mins ago
One Further Question From Student Neighbour.
5 Answers
Unknown Resistor R? dissipates 3.75 Watts. Find the value of R? Ohms. There is no time limit on this one! Once again, student does have an answer. See link below:
https:/ /ibb.co /qB5VK7 w
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Answers
// There are 2 answers after all!! // Let the current through R? be i1 and that passing through the 5 Ω be i2. Then the sum of i1 and i2 equals 2Amps -----> i1 + i2 = 2 Let the voltage across parallel resistor network be V, so that V = 5.i2 Since P = VI then 3.75 = 5.i2.i1 Substituting i1 for 2 - i2 we can write equation as; 3.75 = 5.i2 (2 - i2) Expanding parenthesis and...
09:54 Fri 03rd Mar 2023
>19.19 and >23.03
Incredible!! There are 2 answers after all!!
Yes, both values of R? (1.667 and 15 ohms) divide
the total current (2 Amps) into the ratio of 1 : 3
Sorry guys, but can I ask you to provide an explanation on how you arrived at your answers. Again, many thanks on behalf of my student neighbour and myself!!
Now off to take my daughter to the airport. Who would be a Dad hey LOL?? Catch you all later :-)
Incredible!! There are 2 answers after all!!
Yes, both values of R? (1.667 and 15 ohms) divide
the total current (2 Amps) into the ratio of 1 : 3
Sorry guys, but can I ask you to provide an explanation on how you arrived at your answers. Again, many thanks on behalf of my student neighbour and myself!!
Now off to take my daughter to the airport. Who would be a Dad hey LOL?? Catch you all later :-)
// There are 2 answers after all!! //
Let the current through R? be i1 and that passing through the 5 Ω be i2. Then the sum of i1 and i2 equals 2Amps -----> i1 + i2 = 2
Let the voltage across parallel resistor network be V, so that V = 5.i2
Since P = VI then 3.75 = 5.i2.i1
Substituting i1 for 2 - i2 we can write equation as;
3.75 = 5.i2 (2 - i2)
Expanding parenthesis and rearranging gives;
5.i2² - 10.i2 + 3.75 = 0
The expression above is in the form of a Quadratic equation with roots equating to two values for i2, namely 1.5 and 0.5 Amps.
I will leave that bit upto the HND student to verify.
// If R? = 15Ω then the 5Ω resistor is going to be dissipating 11.25 Watts. //
In terms of achieving minimum power consumption this cannot be disputed. In that sense I totally agree :0)
The value R? = 15 Ω was offered as a pragmatic solution based upon the number 1.5 (coincidentally this happens to be one of the branch current values) and its associated exponents to the powers of 10. These are a preferred resistor value and readily bought off the shelf.
Whereas you will appreciate a 1.667 Ω resistor would be quite difficult to source :-/
Let the current through R? be i1 and that passing through the 5 Ω be i2. Then the sum of i1 and i2 equals 2Amps -----> i1 + i2 = 2
Let the voltage across parallel resistor network be V, so that V = 5.i2
Since P = VI then 3.75 = 5.i2.i1
Substituting i1 for 2 - i2 we can write equation as;
3.75 = 5.i2 (2 - i2)
Expanding parenthesis and rearranging gives;
5.i2² - 10.i2 + 3.75 = 0
The expression above is in the form of a Quadratic equation with roots equating to two values for i2, namely 1.5 and 0.5 Amps.
I will leave that bit upto the HND student to verify.
// If R? = 15Ω then the 5Ω resistor is going to be dissipating 11.25 Watts. //
In terms of achieving minimum power consumption this cannot be disputed. In that sense I totally agree :0)
The value R? = 15 Ω was offered as a pragmatic solution based upon the number 1.5 (coincidentally this happens to be one of the branch current values) and its associated exponents to the powers of 10. These are a preferred resistor value and readily bought off the shelf.
Whereas you will appreciate a 1.667 Ω resistor would be quite difficult to source :-/
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