Transformer Power Equation

Where:

• $V_p$ = potential difference across the primary coil (in Volts, V)
• $I_p$ = current in the primary coil (in Amps, A)
• $V_s$ = potential difference across the secondary coil (in Volts, V)
• $I_s$ = current in the secondary coil (in Amps, A)

The Inverse Relationship

• Because the total power remains constant across a 100% efficient transformer, potential difference and current must be inversely proportional.
• In a step-up transformer, the potential difference increases ($V_s > V_p$), so the current must decrease proportionally ($I_s < I_p$).
• In a step-down transformer, the potential difference decreases ($V_s < V_p$), so the current must increase ($I_s > I_p$).
• This inverse relationship is exactly why the National Grid transmits electricity at extremely high potential differences. Stepping up the potential difference means a much lower current, which drastically reduces the amount of energy wasted as heat in transmission wires.
• Note that this process only works because transformers are supplied with an alternating current (AC), which creates the changing magnetic field needed to induce a potential difference.

Worked Example: Calculating Current

A step-up transformer at a power station is assumed to be 100% efficient. It takes an input potential difference of $20,000\text{ V}$ and steps it up to $400,000\text{ V}$ for transmission. If the current in the primary coil is $5,000\text{ A}$, calculate the current in the secondary coil.

Step 1: State the formula and identify the known values.

• $V_p \times I_p = V_s \times I_s$
• $V_p = 20,000\text{ V}$, $V_s = 400,000\text{ V}$,

Step 2: Substitute the values into the equation.

• $20,000 \times 5,000 = 400,000 \times I_s$

Step 3: Rearrange to solve for $I_s$.

• $100,000,000 = 400,000 \times I_s$
• $I_s = \frac{100,000,000}{400,000}$

Step 4: Calculate the final answer with units.

• $I_s = 250\text{ A}$

Worked Example: Calculating Potential Difference

A laptop charger uses a step-down transformer to convert a $230\text{ V}$ mains a.c. supply into a safe output for the battery. The primary current is $0.15\text{ A}$ and the secondary current is $2.3\text{ A}$. Assuming 100% efficiency, calculate the potential difference across the secondary coil.

Step 1: State the formula and identify the known values.

• $V_p \times I_p = V_s \times I_s$
• $V_p = 230\text{ V}$, $I_p = 0.15\text{ A}$,

Step 2: Substitute the values into the equation.

• $230 \times 0.15 = V_s \times 2.3$

Step 3: Rearrange to solve for $V_s$.

• $34.5 = V_s \times 2.3$
• $V_s = \frac{34.5}{2.3}$

Step 4: Calculate the final answer with units.

• $V_s = 15\text{ V}$