Calculating Voltage, Current, Power, and Energy in Basic Circuits

Welcome to the first lesson of the course. Before designing MOS amplifiers, we need a dependable language for describing any circuit: voltage polarities, current-reference directions, and the sign of power. These are not merely introductory conventions. The same bookkeeping will later let you determine whether a MOSFET, current mirror, or supply is consuming power or delivering it.

In this lesson you will use Ohm’s law to relate voltage, current, and resistance; apply the passive sign convention to interpret signs consistently; and calculate power and energy in basic DC circuits.

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Voltage, current, and resistance: establish reference directions first

A voltage is always a difference between two terminals. If an element’s upper terminal is labelled $+$ and its lower terminal $-$, its defined voltage is

$$v=V_{+}-V_{-}$$

Thus, $v=+2\text{V}$ means the $+$ terminal is 2 V above the $-$ terminal. If calculation gives $v=-2\text{V}$, the physical terminal marked $+$ is actually 2 V below the one marked $-$. The label is a chosen reference, not a promise about the eventual sign.

Likewise, a current arrow is a reference direction. A positive result means that current flows in the arrow’s direction; a negative result means that the actual conventional current flows opposite to the arrow.

For an ideal resistor of resistance $R$, Ohm’s law is

$$v=iR$$

provided $v$ and $i$ are labelled according to the passive sign convention discussed next.

Useful units and prefixes:

Quantity	Symbol	SI unit	Common IC-scale values
Voltage	$v$	volt, V	mV, V
Current	$i$	ampere, A	$\mu \text{A}$, mA
Resistance	$R$	ohm, $\Omega$	$\Omega$, $\text{k}\Omega$, $\text{M}\Omega$
Power	$p$	watt, W	$\mu \text{W}$, mW
Energy	$w$	joule, J	pJ, nJ, $\mu \text{J}$

The unit relationship in Ohm’s law is also useful as a quick check:

$$1\text{V}=1\text{A}\cdot \Omega$$

For example, if a $2\text{k}\Omega$ resistor has $4\text{V}$ across it, with the current reference entering the terminal labelled $+$,

$$i=\frac{v}{R}=\frac{4\text{V}}{2\text{k}\Omega }=2\text{mA}$$

The positive result confirms that the actual current travels in the chosen reference direction.

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The passive sign convention

The passive sign convention (PSC) is the standard arrangement for a two-terminal element:

• label one terminal $+$ and the other $-$;
• draw the positive current reference entering the $+$ terminal;
• define the element power as

$$p=vi$$

Under this convention, the sign of $p$ has a direct physical meaning:

Result from $p=vi$	Interpretation
$p>0$	The element absorbs power. Energy enters it.
$p<0$	The element supplies power. Energy leaves it.
$p=0$	No instantaneous power transfer occurs.

The phrase “passive sign convention” can be slightly misleading at first. It does not say that only resistors, capacitors, and inductors may be labelled this way. You may apply the same convention to a voltage source or current source when calculating power. It is simply a consistent accounting convention.

Why the sign matters

Suppose a $1\text{k}\Omega$ resistor is labelled with $+$ at the top, $-$ at the bottom, and a current reference entering the top. If its voltage is $v=+3\text{V}$, then

$$i=\frac{v}{R}=\frac{3\text{V}}{1\text{k}\Omega }=3\text{mA}$$

Its power is

$$p=vi=(3\text{V})(3\text{mA})=9\text{mW}$$

The resistor absorbs $9\text{mW}$, which in an ordinary resistor becomes heat.

Now keep exactly the same labels but suppose the surrounding circuit imposes $i=-2\text{mA}$. Ohm’s law gives

$$v=iR=(-2\text{mA})(1\text{k}\Omega )=-2\text{V}$$

Both $v$ and $i$ are negative because the actual voltage polarity and current direction are opposite to the labels. Yet the resistor’s power remains positive:

$$p=vi=(-2\text{V})(-2\text{mA})=4\text{mW}$$

This must be so for an ideal positive resistance:

$$p_R=vi=(iR)i=i^{2}R$$

Since $i^2$ and $R$ are nonnegative, a resistor always absorbs or dissipates power. Changing reference labels may change the signs of $v$ and $i$, but it cannot turn a resistor into an energy source.

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Power formulas and how to choose one

Power is the rate at which energy is transferred:

$$p=\frac{dw}{dt}$$

For a two-terminal circuit element using the passive sign convention,

$$p=vi$$

For a resistor, combining $p=vi$ with Ohm’s law yields two alternatives:

$$p=i^{2}R$$ $$p=\frac{v^2}{R}$$

All three expressions give the same result for a resistor. Choose the form that uses the quantities you already know:

Known quantities	Convenient expression
Voltage and current	$p=vi$
Current and resistance	$p=i^{2}R$
Voltage and resistance	$p=v^2/R$

Be careful with a common error: $p=v^2/R$ and $p=i^{2}R$ are resistor-specific because they use Ohm’s law. For a battery, current source, MOSFET, capacitor, or arbitrary circuit block, begin with the universal relation $p=vi$.

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A complete source-and-resistor power calculation

Consider an ideal $5\text{V}$ supply connected across a $1\text{k}\Omega$ resistor. The resistor has the full supply voltage across it.

First calculate the circuit current:

$$i=\frac{5\text{V}}{1\text{k}\Omega }=5\text{mA}$$

For the resistor, current enters the higher-voltage terminal. With the passive sign convention,

$$p_R=v_R{i}_R=(5\text{V})(5\text{mA})=25\text{mW}$$

So the resistor absorbs $25\text{mW}$.

Now label the source using the passive sign convention too: define its voltage as $v_S=+5\text{V}$, with the current reference entering its positive terminal. Physically, the source is driving current out of that terminal, so its signed current is

$$i_S=-5\text{mA}$$

Therefore,

$$p_S=v_S{i}_S=(5\text{V})(-5\text{mA})=-25\text{mW}$$

The negative sign tells us that the source supplies $25\text{mW}$.

The power balance is

$$p_R+p_S=25\text{mW}-25\text{mW}=0$$

This is a compact statement of conservation of energy: in any complete lumped circuit, the algebraic sum of element powers is zero.

$$\sum _k{p}_k=0$$

Some elements absorb energy while others supply it. The signs handle the distinction automatically.

Sources can absorb power too

A source is not required to deliver power at every instant. A rechargeable battery during charging is a familiar example: current enters its positive terminal, so $p=vi>0$. It is then absorbing electrical power and storing energy chemically.

This is why “source” describes the element’s ideal voltage or current behavior, not necessarily the instantaneous direction of energy transfer.

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Energy: power accumulated over time

Power answers, “How rapidly is energy transferred right now?” Energy answers, “How much was transferred over an interval?”

In general,

$$w(t_2)-w(t_1)={\int }_{t_1}^{t_2}p(t)dt$$

When voltage and current are constant, power is constant, and the calculation simplifies to

$$w=pt$$

For the $5\text{V}$, $1\text{k}\Omega$ example, the resistor absorbs $25\text{mW}$. If the circuit operates for $10\text{ms}$,

$$w_R=(25\text{mW})(10\text{ms})=250\mu \text{J}$$

The supply provides the same magnitude of energy:

$$w_S=-250\mu \text{J}$$

The signed energy balance remains zero.

At the IC scale, these numbers matter. A bias branch carrying $100\mu \text{A}$ from a $1.8\text{V}$ supply consumes

$$p=(1.8\text{V})(100\mu \text{A})=180\mu \text{W}$$

If it remains on for $1\text{s}$, it consumes

$$w=(180\mu \text{W})(1\text{s})=180\mu \text{J}$$

Later, when you bias amplifiers, total static power will largely be the sum of supply voltage times the currents drawn from the supply rails. The same PSC calculation will identify the power consumed by each branch.

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A reliable workflow for basic circuit quantities

When solving a basic DC problem, use this order:

1. Choose voltage polarity and current-reference direction for each element. For resistors, use PSC: current enters the labelled $+$ terminal.
2. Write the device relation. For a resistor, use $v=iR$.
3. Solve algebraically before interpreting signs. Do not reverse signs mentally midway through a calculation.
4. Calculate signed power with $p=vi$.
5. Interpret the result: positive means absorbed, negative means supplied.
6. Check power balance when the circuit is complete.
7. Calculate energy using $w=pt$ only when power is constant; otherwise integrate $p(t)$.

Two checks catch many errors:

• A physical resistor with $R>0$ should satisfy $p_R=i^{2}R\ge 0$.
• In a complete circuit, total absorbed power must equal total supplied power in magnitude.

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Key takeaways

Ohm’s law for a resistor is

$$v=iR$$

The passive sign convention defines positive current as entering the terminal labelled positive for voltage. With that convention,

$$p=vi$$

has an immediate interpretation: $p>0$ means absorbed power and $p<0$ means supplied power. For a resistor, the equivalent expressions

$$p=vi=i^{2}R=\frac{v^2}{R}$$

show that an ideal resistor always dissipates nonnegative power.

Finally, power is a rate, while energy is accumulated power:

$$w=\int p(t)dt$$

and, for constant power,

$$w=pt$$

Next, you will use these sign conventions while applying Kirchhoff’s current law and Kirchhoff’s voltage law to solve one-node, two-node, and loop circuits.