HNC Engineering Maths - Vector Diagrams
Hi there,
I've been working out the below question for a while and as soon as I thought I had the right answer I started to question myself.
a) A resistor, R, is connected in series with an inductor, L. An a.c. current, 𝑖, flows through this RL combination, causing a voltage (VR) of 30V to be developed across the resistor, and a voltage (𝑉𝐿) of 40V to be developed across the inductor.
Assuming that VR is in-phase with 𝑖, and VL leads VR by 90°, draw a vector diagram for this arrangement and then calculate the magnitude of the resultant voltage across the whole RL combination.
I have drawn out the vector diagram with VL on the y-axis and VR on the x-axis and used Pythagorus to work out the hypotenuse shown below.
c=√(30^2+40^2 )=√(900+1600)=√2500=50V
However, I haven't taken into account VR being in-phase and VL leading by 90°, and the answer was far to easy so I'm questioning my answer now, can anyone help?
I've been working out the below question for a while and as soon as I thought I had the right answer I started to question myself.
a) A resistor, R, is connected in series with an inductor, L. An a.c. current, 𝑖, flows through this RL combination, causing a voltage (VR) of 30V to be developed across the resistor, and a voltage (𝑉𝐿) of 40V to be developed across the inductor.
Assuming that VR is in-phase with 𝑖, and VL leads VR by 90°, draw a vector diagram for this arrangement and then calculate the magnitude of the resultant voltage across the whole RL combination.
I have drawn out the vector diagram with VL on the y-axis and VR on the x-axis and used Pythagorus to work out the hypotenuse shown below.
c=√(30^2+40^2 )=√(900+1600)=√2500=50V
However, I haven't taken into account VR being in-phase and VL leading by 90°, and the answer was far to easy so I'm questioning my answer now, can anyone help?
Original post by MRL90
Hi there,
I've been working out the below question for a while and as soon as I thought I had the right answer I started to question myself.
a) A resistor, R, is connected in series with an inductor, L. An a.c. current, 𝑖, flows through this RL combination, causing a voltage (VR) of 30V to be developed across the resistor, and a voltage (𝑉𝐿) of 40V to be developed across the inductor.
Assuming that VR is in-phase with 𝑖, and VL leads VR by 90°, draw a vector diagram for this arrangement and then calculate the magnitude of the resultant voltage across the whole RL combination.
I have drawn out the vector diagram with VL on the y-axis and VR on the x-axis and used Pythagorus to work out the hypotenuse shown below.
c=√(30^2+40^2 )=√(900+1600)=√2500=50V
However, I haven't taken into account VR being in-phase and VL leading by 90°, and the answer was far to easy so I'm questioning my answer now, can anyone help?
I've been working out the below question for a while and as soon as I thought I had the right answer I started to question myself.
a) A resistor, R, is connected in series with an inductor, L. An a.c. current, 𝑖, flows through this RL combination, causing a voltage (VR) of 30V to be developed across the resistor, and a voltage (𝑉𝐿) of 40V to be developed across the inductor.
Assuming that VR is in-phase with 𝑖, and VL leads VR by 90°, draw a vector diagram for this arrangement and then calculate the magnitude of the resultant voltage across the whole RL combination.
I have drawn out the vector diagram with VL on the y-axis and VR on the x-axis and used Pythagorus to work out the hypotenuse shown below.
c=√(30^2+40^2 )=√(900+1600)=√2500=50V
However, I haven't taken into account VR being in-phase and VL leading by 90°, and the answer was far to easy so I'm questioning my answer now, can anyone help?
Original post by MRL90
Hi there,
I've been working out the below question for a while and as soon as I thought I had the right answer I started to question myself.
a) A resistor, R, is connected in series with an inductor, L. An a.c. current, 𝑖, flows through this RL combination, causing a voltage (VR) of 30V to be developed across the resistor, and a voltage (𝑉𝐿) of 40V to be developed across the inductor.
Assuming that VR is in-phase with 𝑖, and VL leads VR by 90°, draw a vector diagram for this arrangement and then calculate the magnitude of the resultant voltage across the whole RL combination.
I have drawn out the vector diagram with VL on the y-axis and VR on the x-axis and used Pythagorus to work out the hypotenuse shown below.
c=√(30^2+40^2 )=√(900+1600)=√2500=50V
However, I haven't taken into account VR being in-phase and VL leading by 90°, and the answer was far to easy so I'm questioning my answer now, can anyone help?
I've been working out the below question for a while and as soon as I thought I had the right answer I started to question myself.
a) A resistor, R, is connected in series with an inductor, L. An a.c. current, 𝑖, flows through this RL combination, causing a voltage (VR) of 30V to be developed across the resistor, and a voltage (𝑉𝐿) of 40V to be developed across the inductor.
Assuming that VR is in-phase with 𝑖, and VL leads VR by 90°, draw a vector diagram for this arrangement and then calculate the magnitude of the resultant voltage across the whole RL combination.
I have drawn out the vector diagram with VL on the y-axis and VR on the x-axis and used Pythagorus to work out the hypotenuse shown below.
c=√(30^2+40^2 )=√(900+1600)=√2500=50V
However, I haven't taken into account VR being in-phase and VL leading by 90°, and the answer was far to easy so I'm questioning my answer now, can anyone help?
Original post by MRL90
For context here is the vector diagram
Your work has considered the phase difference of VL and VR of 90° when you use Pythagorean theorem to find the resultant voltage of VL and VR.
You can find more info here.
Original post by Eimmanuel
Your work has considered the phase difference of VL and VR of 90° when you use Pythagorean theorem to find the resultant voltage of VL and VR.
You can find more info here.
You can find more info here.
Thanks for this! I've had a hard time trying to find information online, so it would appear that I have the right answer and I'm just overthinking about it being too easy 😂
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