Physics 13.10 - Ideal transformer power

Physics 13.10 - Ideal transformer power

Use equal input and output power to calculate transformer voltages and currents. Work with the 100%-efficient model, convert units consistently, and check whether a proposed output obeys energy conservation. This content is assessed at both Foundation and Higher tiers.

Equal input and output power

The primary is the input side of a transformer and the secondary is the output side. A transformer changes voltage but is not an energy source. For the ideal model of 100% efficiency, no energy is dissipated inside the transformer, so input power equals output power.

Electrical power is potential difference multiplied by current. Therefore:

An ideal transformer

VpIp=VsIsV_{\mathrm p} I_{\mathrm p}=V_{\mathrm s} I_{\mathrm s}
SymbolQuantityUnit
VpV_{\mathrm p}Potential difference across the primaryvolt, V\mathrm{V}
IpI_{\mathrm p}Current in the primaryampere, A\mathrm{A}
VsV_{\mathrm s}Potential difference across the secondaryvolt, V\mathrm{V}
IsI_{\mathrm s}Current in the secondaryampere, A\mathrm{A}

The product on each side is power in watts. In a step-up transformer, voltage increases and current decreases; a step-down transformer decreases voltage and increases current. These comparisons assume the same ideal transfer of power, not a power gain. Real transformers dissipate some energy, so the equality is an approximation only when 100% efficiency is specified.

Calculating secondary current

Worked example: finding secondary current

An ideal transformer has Vp=25,000 VV_{\mathrm p}=25{,}000\,\mathrm{V}, Ip=80 AI_{\mathrm p}=80\,\mathrm{A} and Vs=400 kVV_{\mathrm s}=400\,\mathrm{kV}.

First, 400 kV=400,000 V400\,\mathrm{kV}=400{,}000\,\mathrm{V}. Rearrange:

Is=VpIpVsI_{\mathrm s}=\frac{V_{\mathrm p}I_{\mathrm p}}{V_{\mathrm s}} Is=25,000×80400,000=5.0 AI_{\mathrm s}=\frac{25{,}000\times80}{400{,}000}=5.0\,\mathrm{A}

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