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184_notes:examples:week14_step_down_transformer [2017/11/28 21:53] – created tallpaul | 184_notes:examples:week14_step_down_transformer [2021/07/22 13:56] (current) – [Solution] schram45 | ||
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+ | [[184_notes: | ||
+ | |||
===== Designing a Step-down Transformer ===== | ===== Designing a Step-down Transformer ===== | ||
+ | Recall the [[184_notes: | ||
+ | |||
+ | ===Facts=== | ||
+ | * The high-voltage line carries $240 \text{ kV}$. | ||
+ | * We want the low-voltage line to carry $120 \text{ V}$. | ||
+ | * We know the step-up transformer was created by wrapping two solenoids around an iron ring, and connected the high-voltage line to one solenoid, and the low-voltage line to the other. The low-voltage solenoid had fewer turns. A more detailed representation is shown below. | ||
+ | |||
+ | ===Lacking=== | ||
+ | * We need a design for the step-down transformer. | ||
+ | |||
+ | ===Approximations & Assumptions=== | ||
+ | * We have access to the same materials as we did for the step-up transformer: | ||
+ | * The step-down transformer we are building will have a similar design to the step-up transformer. | ||
+ | |||
+ | ===Representations=== | ||
+ | * We represent magnetic flux as $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$$ | ||
+ | * We can represent induced voltage as $$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t}$$ | ||
+ | * We represent the step-up transformer, | ||
+ | |||
+ | [{{ 184_notes: | ||
+ | ====Solution==== | ||
+ | It makes sense intuitively to just flip the step-up transformer for our design. If the step-up transformer brings voltage down, we could just reverse it to bring the voltage up. Our design is shown below, where **now the number of turns in the primary solenoid is much greater than the number of turns in the secondary solenoid** -- it's just the flipped step-up transformer, | ||
+ | |||
+ | [{{ 184_notes: | ||
+ | |||
+ | As with the step-up transformer, | ||
+ | |||
+ | [{{ 184_notes: | ||
+ | |||
+ | The iron is able to align its atoms with the magnetic field much faster than the current alternates between directions, which is why we draw the magnetic field the same everywhere. The iron also greatly amplifies the magnetic field that the primary solenoid would produce in air, so even though $B_P$ contains magnetic field contributions from the primary solenoid //and// from the iron, the contribution from the iron is far greater. For this reason, we approximate the magnetic field as the same at all locations in the iron. By this approximation, | ||
+ | |||
+ | Since the magnetic field in the iron ring is changing with the alternating current, (in the primary solenoid), this will induce a voltage ($V_S$) in the secondary solenoid. We can use Faraday' | ||
+ | $$-V_{S}=\frac{d\Phi_{B_{S}}}{dt}$$ | ||
+ | |||
+ | The rest of our calculation follows just as it would for the [[184_notes: | ||
+ | $$\Phi_{B_{S}}=\int \vec{B}_S \bullet \vec{dA}_S$$ | ||
+ | where the area here would be the cross-sectional area of the iron ring (since this is where the magnetic field is). The direction of the magnetic field would always be perpendicular to the area of the cylinder - so the flux would simplfy to: | ||
+ | $$\Phi_{B_{S}}=B_S A_S$$ | ||
+ | But this is the magnetic flux through one loop in the secondary solenoid - if we want the total flux through all the loops then we have to multiply by the number of loops: | ||
+ | $$\Phi_{B_{S}}=B_S A_S N_S$$ | ||
+ | We can then plug this into the voltage equation we wrote above: | ||
+ | $$-V_{S}=\frac{d}{dt}\biggl(B_S A_S N_S\biggr)$$ | ||
+ | |||
+ | Since $N_S$ and $A_S$ are constant with respect to time (not adding/ | ||
+ | $$-V_{S}=N_S A_S \frac{d}{dt}\biggl(B_S\biggr)$$ | ||
+ | //__If we assume that the iron ring has a constant cross-sectional area__//, then $A_P=A_S$ and we already said that $B_P=B_S$. This means we can rewrite the flux through the secondary solenoid in terms of the magnetic field and area of the primary solenoid. | ||
+ | $$-V_{S}=N_S A_P \frac{d}{dt}\biggl(B_P\biggr)$$ | ||
+ | |||
+ | If we look at the $A_P\frac{d}{dt}\biggl(B_P\biggr)$ term, this looks very much like the changing magnetic flux (or induced voltage) through a single loop of the //primary// solenoid: | ||
+ | $$\frac{d\Phi_{B_{P}}}{dt}=\frac{d}{dt}\biggl(B_P A_P\biggr)$$ | ||
+ | If we use Faraday' | ||
+ | $$-V_P=N_P \frac{d}{dt}\biggl(B_P A_P\biggr)$$ | ||
+ | $$\frac{d}{dt}\biggl(B_P A_P\biggr)=\frac{-V_P}{N_P}$$ | ||
+ | |||
+ | Finally, We can plug this result into the $V_S$ equation above to get the potential in the secondary solenoid as it relates to the potential in the primary solenoid: | ||
+ | $$-V_S = N_S \frac{-V_P}{N_P}$$ | ||
+ | $$V_S=V_P \frac{N_S}{N_P}$$ | ||
+ | |||
+ | Remember, we need to step down from the $240 \text{ kV}$ power line to a $120 \text{ V}$ line. This is a factor of 2000. One way to achieve this would be to set $N_P = 20000$, and $N_S = 10$. Due to the huge step down, it may be even easier to design a series of step-down transformers, |